Past JEE Main Entrance Papers Set-I

Hint

\(
\begin{aligned}
& \vec{l}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \\
& \vec{l}_2=4 \hat{i}+6 \hat{j}+8 \hat{k}
\end{aligned}
\)
\(
S . D .=\frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times((2-\lambda) \hat{i}-4 \hat{k})|}{|2 \hat{i}+3 \hat{j}+4 \hat{k}|}
\)
\(
\begin{aligned}
& \left\lvert\, \frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times((2-\lambda) \hat{i}+4 \hat{k})|}{\sqrt{29}}=\frac{13}{\sqrt{29}}\right. \\
& |-8 \hat{j}-3(2-\lambda) \hat{k}+12 \hat{i}+4(2-\lambda) \hat{j}|=13 \\
& |12 \hat{i}-4 \lambda \hat{j}+(3 \lambda-6) \hat{k}|=13 \\
& 144+16 \lambda^2+(3 \lambda-6)^2=169 \\
& 16 \lambda^2+(3 \lambda-6)^2=25=\lambda \Rightarrow=1
\end{aligned}
\)

Hint

\(
L: \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\mu
\)
Measured along \(L_2: \frac{x-\frac{11}{3}}{\frac{2}{3}}=\frac{y-\frac{11}{3}}{\frac{1}{3}}=\frac{z-\frac{19}{3}}{\frac{2}{3}}=\lambda\)
Any point on \(L_1:(\mu+1, \mu+2,2 \mu+3)\)
Any point on \(L_2\left(\frac{2}{3} \lambda+\frac{11}{3}, \frac{\lambda}{3}+\frac{11}{3}, \frac{2}{3} \lambda+\frac{19}{3}\right)\)
Now
\(
\begin{aligned}
& \mu+1=\frac{2}{3} \lambda+\frac{11}{3} \\
& \frac{\mu+2=\frac{\lambda}{3}+\frac{11}{3}}{\lambda=-3} \\
& \mu=\frac{2}{3}
\end{aligned}
\)
Point on \(L=\left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right)\)
\(
\begin{aligned}
& d=\sqrt{\left(\frac{11}{3}-\frac{5}{3}\right)^2+\left(\frac{8}{3}-\frac{11}{3}\right)^2+\left(\frac{19}{3}-\frac{13}{3}\right)^2} \\
& d=\sqrt{4+1+4} \\
& d=3
\end{aligned}
\)

Hint

Given lines are
\(
\begin{aligned}
& \frac{x-3}{4}=\frac{y-(-7)}{-11}=\frac{z-1}{5} \text { and } \\
& \frac{x-5}{3}=\frac{y-9}{-6}=\frac{z-(-2)}{1}
\end{aligned}
\)
Shortest distance between two lines,
\(
\begin{aligned}
& d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|} \\
& \vec{a}_2-\vec{a}_1=2 \hat{i}+16 \hat{j}-3 \hat{k} \text { and } \\
& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
4 & -11 & 5 \\
3 & -6 & 1
\end{array}\right|=19 \hat{i}+11 \hat{j}+9 \hat{k} \\
& \therefore \quad d=\frac{187}{\sqrt{563}}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& L_1: \frac{x-2}{1}=\frac{y}{-1}=\frac{z-1}{1}=\lambda \\
& L_2: \frac{x+1}{(1 / 2)}=\frac{y-1}{(1 / 2)}=\frac{z+1}{1}=\mu
\end{aligned}
\)
Any point of \(L_1\) and \(L_2\) will be \((\lambda+2,-\lambda, \lambda+1)\) and \(\left(\frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1\right)\)
Now DR of line \(<\lambda-\frac{\mu}{2}+3,-\lambda-\frac{\mu}{2}-1, \lambda-\mu+2>\)
Now \(\frac{\lambda-\frac{\mu}{3}+3}{3}=\frac{-\lambda-\frac{\mu}{2}-1}{1}=\frac{\lambda-\mu+2}{2}\)
\(
\left.\begin{array}{c}
\lambda-\frac{\mu}{3}+3=-3 \lambda-\frac{3 \mu}{2}-3 \ldots(1) \\
2\left(\lambda-\frac{\mu}{3}+3\right)=3(\lambda-\mu+2) \ldots(2)
\end{array}\right\} \lambda=\frac{-4}{3}, \mu=\frac{-2}{3}
\)
\(\begin{aligned} & \text { (1) } \\ & \text { (2) }\end{aligned} \lambda \lambda=\frac{-4}{3}, \mu=\frac{-2}{3}\)
\(\therefore\) Points will be \(\left(\frac{2}{3}, \frac{4}{3}, \frac{-1}{3}\right)\) and \(\left(\frac{-4}{3}, \frac{2}{3}, \frac{-5}{3}\right)\)
\(\therefore \quad L\) will be \(\frac{x-\frac{2}{3}}{3}=\frac{y-\frac{4}{3}}{1}=\frac{z+\frac{1}{3}}{2}\)
\(\therefore \quad\left(\frac{-1}{3}, 1,-1\right)\) will satisfy \(L\)

Hint

\(
\begin{aligned}
& \overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k} \\
& \overrightarrow{O Q}=x \hat{i}+y \hat{j} \\
& |O P|=\gamma=\sqrt{x^2+y^2+z^2} \\
& \cos \theta=\frac{x}{\sqrt{x^2+y^2}} \Rightarrow \cos ^2 \theta=\frac{x^2}{\gamma^2-z^2}=\frac{x^2}{\gamma^2-\gamma^2 \cos ^2 \phi} \\
& \cos \phi=\frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{z}{\gamma}
\end{aligned}
\)
Distance of \(P\) from \(x\)-axis \(=\sqrt{y^2+z^2}\)
\(
\begin{aligned}
& d=\sqrt{\gamma^2-x^2} \\
& \Rightarrow x^2=\gamma^2 \sin ^2 \phi \cos ^2 \theta \\
& \Rightarrow d=\sqrt{\gamma^2-\gamma^2 \sin ^2 \phi \cos ^2 \theta} \\
& =\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k} \\
& L_1=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+4 \hat{k}) \\
& L_2: \vec{r}=2 \hat{i}+3 \hat{j}+5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k}) \\
& \vec{a}_1=2 \hat{i}+\hat{j}+3 \hat{k} \\
& \vec{a}_2=2 \hat{i}+3 \hat{j}+5 \hat{k} \\
& \vec{a}_2-\vec{a}_1=2 \hat{j}+2 \hat{k} \\
& \vec{b}_1=\hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}_2=2 \hat{i}+3 \hat{j}+\hat{k} \\
& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -3 & 4 \\
2 & 3 & 1
\end{array}\right|
\end{aligned}
\)
\(
\begin{aligned}
& \hat{i}(-3-12)-\hat{j}(1-8)+\hat{k}(3+6) \\
& =-15 \hat{i}+7 \hat{j}+9 \hat{k} \\
& \left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{225+49+81} \\
& \left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\frac{14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}} \\
& m+n=387
\end{aligned}
\)

Hint

\(
\begin{aligned}
& RQ =\sqrt{1+64+4}=\sqrt{69} \\
& \overrightarrow{ RQ }=\hat{\ell}-8 \hat{ j }+2 \hat{ k } \\
& \overrightarrow{ RS }=\hat{\ell}+\hat{ j }-\hat{ k } \\
& \cos \theta=\frac{\overrightarrow{ RQ } \cdot \overrightarrow{ RS }}{|\overrightarrow{ RQ }||\overrightarrow{ RS }|}=\left|\frac{1-8-2}{\sqrt{69} \sqrt{3}}\right|=\frac{9}{3 \sqrt{23}} \\
& \cos \theta=\frac{3}{\sqrt{23}}=\frac{ RS }{ RQ }=\frac{ RS }{\sqrt{69}} \\
& RS =3 \sqrt{3} \\
& \sin \theta=\frac{\sqrt{14}}{\sqrt{23}}=\frac{ QS }{\sqrt{69}} \\
& QS =\sqrt{42} \\
& \operatorname{area}=\frac{1}{2} \cdot 2 QS \cdot RS =\sqrt{42} \cdot 3 \sqrt{3} \\
& \lambda=9 \sqrt{14} \\
& \lambda^2=81.14=14 k \\
& k =81
\end{aligned}
\)

Hint

\(A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1), D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)\) are vertices of a quadrilateral.
\(
\begin{aligned}
& \overrightarrow{A C}=(2 \hat{i}+2 \hat{j}+\hat{k})-(3 \hat{i}+\hat{j}-\hat{k}) \\
& =-\hat{i}+\hat{j}+2 \hat{k} \\
& \overrightarrow{B D}=\left(\frac{10}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}\right)-\left(\frac{5}{3} \hat{i}+\frac{7}{3} \hat{j}+\frac{1}{3} \hat{k}\right) \\
& \overrightarrow{B D}=\frac{5}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{2}{3} \hat{k} \\
& \text { Area }=\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B D}| \\
& =\frac{1}{2}\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 1 & 2 \\
\frac{5}{3} & \frac{-5}{3} & \frac{-2}{3}
\end{array}\right| \\
& =\frac{1}{2} \sqrt{\left(\frac{8}{3}\right)^2+\left(\frac{8}{3}\right)^2}\left[\because \overrightarrow{A C} \times \overrightarrow{B D}=\frac{8}{3} \hat{i}+\frac{8}{3} \hat{j}\right] \\
& =\frac{1}{2} \times \frac{8}{3} \times \sqrt{2}=\frac{4 \sqrt{2}}{3} \\
&
\end{aligned}
\)

Hint

Given two lines are \(\frac{x-3}{2}=\frac{y-(-15)}{-7}=\frac{z-9}{5}\) and
\(
\frac{x-(-1)}{2}=\frac{y-1}{1}=\frac{z-9}{-3}
\)
Shortest distance between two lines are
\(
\begin{aligned}
& d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\vec{b}_1 \times \vec{b}_2\right|} \\
& \therefore \quad \vec{a}_2-\vec{a}_1=-4 \hat{i}+16 \hat{j}+0 \hat{k} \\
& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -7 & 5 \\
2 & 1 & -3
\end{array}\right|=16 \hat{i}+16 \hat{j}+16 \hat{k} \\
& \therefore \quad d=\frac{|-64+16 \times 16|}{16 \sqrt{3}}=4 \sqrt{3}
\end{aligned}
\)

Hint

Let \(Q\) be general point.
\(
(x, y, z)=(2 \lambda+1,3 \lambda-1,5 \lambda+2)
\)
\(\therefore\) Now D.R. of P.Q
\(
\begin{aligned}
& P Q \Rightarrow(2 \lambda+1-8,3 \lambda-1-5,5 \lambda+2-7) \\
& =(2 \lambda-7,3 \lambda-6,5 \lambda-5)
\end{aligned}
\)
DR of line : \((2,3,5)\)
\(
\begin{array}{ll}
\therefore & 2(2 \lambda-7)+3(3 \lambda-6)+5(5 \lambda-5)=0 \\
\Rightarrow & \lambda=\frac{3}{2} \\
\therefore & Q\left(4, \frac{7}{2}, \frac{19}{2}\right) \\
\therefore & (\alpha, \beta, y) \equiv(0,2,12) \quad\left(Q \text { is mid point of } P \& P^{\prime}\right) \\
& \alpha+\beta+y \equiv 14
\end{array}
\)

Hint

\(
\begin{aligned}
& L_1: \frac{x-2}{(-3)}=\frac{y-\frac{2}{3}}{\left(\frac{4 \lambda+1}{3}\right)}=\overline{(-1)} \\
& L_2: \frac{x+3}{3 \mu}=\frac{y-\frac{1}{2}}{-3}=\frac{z-5}{-7} \\
& \because L_1 \perp L_2 \\
& \Rightarrow(-3)(3 \mu)+\left(\frac{4 \lambda+1}{3}\right)(-3)+(-1)(-7)=0 \\
& -9 \mu-4 \lambda-1+7=0 \\
& \Rightarrow 4 \lambda+9 \mu=6
\end{aligned}
\)

Hint

\(
\begin{aligned}
& P_1:(3 k-6,2 k, k-1) \\
& P_2(4 \alpha+7,3 \alpha+9,2 \alpha+4) \\
& P_1 \equiv P_2 \\
& 3 k-6=4 \alpha+7 \Rightarrow 3 k-4 \alpha=13 \\
& 2 k=3 \alpha+9 \Rightarrow 2 k-3 \alpha=9 \\
& \therefore k=3, \alpha=-1 \\
& \therefore P_1:(3,6,2)
\end{aligned}
\)
Distance of \((3,6,2)\) and \((7,8,9)\)
\(
\begin{aligned}
& =\sqrt{16+4+49}=\sqrt{69}=d \\
& d^2+6=69+6=75
\end{aligned}
\)

Hint

\(
\begin{aligned}
& L_1: \frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1} \\
& L_2: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}
\end{aligned}
\)
Point of intersection of \(L_1\) and \(L_2\) is \((-1,1,-1)\)
Distance of point \(P\) from \(L_3: 4 x=2 y=z\)
\(
L_3: \frac{x}{\frac{1}{4}}=\frac{y}{\frac{1}{2}}=\frac{z}{1}
\)
Any point on \(L_3\) be
\(
\begin{aligned}
& \left(\frac{\lambda}{4}, \frac{\lambda}{2}, \lambda\right) \\
& P R:\left\langle\frac{\lambda}{4}+1, \frac{\lambda}{2}-1, \lambda+1\right\rangle \\
& \because P R \perp\left\langle\frac{1}{4}, \frac{1}{2}, 1\right\rangle \\
& \Rightarrow\left(\frac{\lambda}{4}+1\right) \frac{1}{4}+\frac{1}{2}\left(\frac{\lambda}{2}-1\right)+\lambda+1=0 \\
& \frac{\lambda}{16}+\frac{1}{4}+\frac{\lambda}{4}-\frac{1}{2}+\lambda+1=0 \\
& \Rightarrow \quad \lambda=\frac{-4}{7}
\end{aligned}
\)
\(
\begin{aligned}
& \therefore R\left(\frac{-1}{7}, \frac{-2}{7}, \frac{-4}{7}\right) \\
& \text { Now } R P: \sqrt{\left(\frac{-1}{7}+1\right)^2+\left(\frac{-2}{7}-1\right)^2+\left(\frac{-4}{7}+1\right)^2} \\
& =\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7}=\frac{3 \sqrt{14}}{7}
\end{aligned}
\)

Hint

Line through \(P Q\)
\(
\frac{x-1}{4}=\frac{y+2}{-2}=\frac{z-3}{4}
\)
Any point on \(P Q\). be \(R(4 \lambda+1,-2 \lambda-2,4 \lambda+3)\)
\(
P R=9 \text { unit }
\)
\(
\begin{aligned}
& (P R)^2=81 \\
& (4 \lambda+1-1)^2+(-2 \lambda-2+2)^2+(4 \lambda+3-3)^2=81 \\
& 16 \lambda^2+4 \lambda^2+16 \lambda^2=81 \\
& 36 \lambda^2=81 \\
& \lambda= \pm \frac{9}{6}= \pm \frac{3}{2} \\
& \therefore R \text { can be }(7,-5,9) \text { or }(-5,1,-3)
\end{aligned}
\)
Distance from origin for both points be \(\sqrt{49+25+81}\) and \(\sqrt{25+1+9}=\sqrt{35}\)
\(\therefore\) Distance of \((7,-5,9)\) is farthest from origin
\(
\therefore(\alpha, \beta, \gamma)=(7,-5,9)
\)
Now \(7^2+(-5)^2+9^2=155\)

Hint

\(D\) divides \(B C\) in ratio \(1: 1\)
\(
\begin{aligned}
& D:\left(\frac{1}{2}, 7, \frac{7}{2}\right) \\
& \overrightarrow{A D}=\left(\frac{1}{2}-1\right) \hat{i}+(7-3) \hat{j}+\left(\frac{7}{2}-2\right) \hat{k} \\
& =-\frac{1}{2} \hat{i}+4 \hat{j}+\frac{3}{2} \hat{k} \\
& \overrightarrow{A C}=2 \hat{i}+3 \hat{j}+5 \hat{k}
\end{aligned}
\)
Projection of \(\overrightarrow{A D}\) on \(\overrightarrow{A C}\)
\(
=\frac{-1+12+\frac{15}{2}}{\sqrt{4+9+25}}=\frac{37}{2 \sqrt{38}}
\)

Hint

\(
\begin{aligned}
& \text { Any point on line } \frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2} \\
& \text { can be taken as }(8 \lambda-3,2 \lambda+4,2 \lambda-1) \\
& \text { If at a distance of } 6 \text { units from } R(1,2,3) \\
& \Rightarrow(8 \lambda-3-1)^2+(2 \lambda+4-2)^2+(2 \lambda-1-3)^2=36 \\
& \Rightarrow \lambda^2-\lambda=0 \text { \{on simplification \} } \\
& \Rightarrow \lambda=0, \lambda=1
\end{aligned}
\)
Here \(P \& Q\) are \((-3,4,-1)\) and \((5,6,1)\) Centroid of \(\triangle P Q R\)
\(
\begin{aligned}
& (\alpha, \beta, \gamma) \equiv\left(\frac{5-3+1}{3}, \frac{6+4+2}{3}, \frac{1-1+3}{3}\right) \\
& \Rightarrow \alpha=1, \beta=4, \gamma=1 \\
& \Rightarrow \alpha^2+\beta^2+\gamma^2=18
\end{aligned}
\)

Hint

\(
L : \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}=\lambda, p (3,4,9)
\)
\(
\begin{aligned}
& M (3 \lambda+1,2 \lambda-1, \lambda+2) \\
& \overrightarrow{ PM }=(3 \lambda-2,2 \lambda-5, \lambda-7)
\end{aligned}
\)
DR’s of \(L :(3,2,1)\)
\(\overrightarrow{ PM } \perp\) line \(L\)
So, \(3(3 \lambda-2)+2(2 \lambda-5)+1(\lambda-7)=0\)
\(
\begin{aligned}
& 9 \lambda-6+4 \lambda-10+\lambda-7=0 \\
& 14 \lambda=23 \Rightarrow \lambda=\frac{23}{14} \\
& M \left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}\right)
\end{aligned}
\)
Now, As \(M\) is mid-point of \(pp ^{\prime}\)
\(\therefore\) co-ordinates of \(p ^{\prime}\) are
\(
\begin{gathered}
\frac{\alpha+3}{2}=\frac{83}{14} \Rightarrow \alpha=\frac{62}{7} \\
\frac{\beta+4}{2}=\frac{32}{14} \Rightarrow \beta=\frac{4}{7} \\
\frac{\gamma+9}{2}=\frac{51}{14} \Rightarrow \gamma=\frac{-12}{7}
\end{gathered}
\)
So, \(14(\alpha+\beta+\gamma)=14 \times \frac{54}{7} \Rightarrow 108\)

Hint

Given the two lines:
\(
\begin{aligned}
& L_1: \frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1} \\
& L_2: \frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}
\end{aligned}
\)
We observe that these lines are not parallel as their directional vectors are not proportional. The directional vector for \(L_1\) is \((-2,1,1)\) and for \(L_2\) is \((1,-2,1)\). The shortest distance between two skew (non-intersecting and non-parallel) lines in the threedimensional space is along the line that is perpendicular to both lines. This implies we can find a vector that is perpendicular to both directional vectors by taking their cross product.
The directional vector for \(L_1\) is \(d_1=\langle-2,1,1\rangle\), and for \(L_2\) is \(d_2=\langle 1,-2,1\rangle\). The cross product of \(d_1\) and \(d_2\), which will be perpendicular to both lines, is given by:
\(
d=d_1 \times d_2=\left|\begin{array}{ccc}
i & j & k \\
-2 & 1 & 1 \\
1 & -2 & 1
\end{array}\right|
\)
Expanding the determinate gives:
\(
\begin{aligned}
& d= i ((1)(1)-(1)(-2))- j ((-2)(1)-(1)(1))+ k ((-2)(-2)-(1)(1)) \\
& d= i (1+2)- j (-2-1)+ k (4-1) \\
& d=3 i +3 j +3 k
\end{aligned}
\)
\(
d=\langle 3,3,3\rangle
[latex]
The shortest distance [latex]D\) between the two lines can then be given by the formula:
\(
D=\frac{\left|\left( a _2- a _1\right) \cdot d\right|}{\|d\|}
\)
Where \(a _1\) and \(a _{ 2 }\) are position vectors to any points on line \(L_1\) and line \(L_2\), respectively, and ‘.’ denotes the dot product.
From the lines’ equations, we can choose a point on each line (when the parameter is zero). Thus, for \(L_1\), let’s choose the point \(A(\lambda, 2,1)\), and for \(L_2\), let’s choose the point \(B(\sqrt{3}, 1,2)\). These points correspond to the vectors \(a _1=\langle\lambda, 2,1\rangle\) and \(a _2=\langle\sqrt{3}, 1,2\rangle\), respectively.
The vector \(a _2- a _1\) is:
\(
\begin{aligned}
& a _2- a _1=\langle\sqrt{3}, 1,2\rangle-\langle\lambda, 2,1\rangle \\
& a _2- a _1=\langle\sqrt{3}-\lambda, 1-2,2-1\rangle \\
& a _2- a _1=\langle\sqrt{3}-\lambda,-1,1\rangle
\end{aligned}
\)
We can now substitute this, along with \(d\), into the distance formula:
\(
\begin{aligned}
& D=\frac{|((\sqrt{3}-\lambda,-1,1\rangle) \cdot(3,3,3\rangle|}{\|(3,3,3)\|} \\
& D=\frac{|3(\sqrt{3}-\lambda)+3(-1)+3(1)|}{\sqrt{3^2+3^2+3^2}} \\
& D=\frac{|3 \sqrt{3}-3 \lambda-3+3|}{\sqrt{27}} \\
& D=\frac{|3 \sqrt{3}-3 \lambda|}{3 \sqrt{3}} \\
& D=\frac{|\sqrt{3}-\lambda|}{\sqrt{3}}
\end{aligned}
\)
Given that the shortest distance \(D\) between the lines is 1 , we can equate the above result to 1 , and solve for \(\lambda\) :
\(
\begin{aligned}
& \frac{|\sqrt{3}-\lambda|}{\sqrt{3}}=1 \\
& |\sqrt{3}-\lambda|=\sqrt{3}
\end{aligned}
\)
This absolute value equation gives us two possible cases:
Case 1: \(\sqrt{3}-\lambda=\sqrt{3}\), which gives \(\lambda=0\).
Case 2: \(\sqrt{3}-\lambda=-\sqrt{3}\), which gives \(\lambda=2 \sqrt{3}\).
Therefore, the sum of all possible values of \(\lambda\) is:
\(
\lambda_{\text {sum }}=\lambda_1+\lambda_2=0+2 \sqrt{3}=2 \sqrt{3}
\)
Hence, option \(B(2 \sqrt{3})\) is the correct answer.

Hint

\(
\begin{aligned}
& \because \overrightarrow{ PR } \perp(2,3,4) \\
& \therefore \overrightarrow{ PR } \cdot(2,3,4)=0 \\
& (\alpha-2, \beta-3, \gamma-5) \cdot(2,3,4)=0 \\
& \Rightarrow 2 \alpha+3 \beta+4 \gamma=4+9+20=33
\end{aligned}
\)

Hint

\(
L _2=\frac{ x +4}{3}=\frac{ y -4}{2}=\frac{ z -3}{0}
\)
\(
\therefore S . D =\frac{\left|\begin{array}{ccc}
x _2- x _1 & y _2- y _1 & z _2- z _1 \\
2 & -3 & 2 \\
3 & 2 & 0
\end{array}\right|}{\left|\overrightarrow{ n _1} \times \overrightarrow{ n _2}\right|}
\)
\(
=\frac{\left|\begin{array}{ccc}
5 & -5 & -7 \\
2 & -3 & 2 \\
3 & 2 & 0
\end{array}\right|}{\left|\overrightarrow{ n _1} \times \overrightarrow{ n _2}\right|}
\)
\(
\begin{aligned}
& =\frac{141}{|-4 \hat{ i }+6 \hat{ j }+13 \hat{ k }|} \\
& =\frac{141}{\sqrt{16+36+169}} \\
& =\frac{141}{\sqrt{221}}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& L _1 \perp L _2 \quad L _3 \perp L _1, L _2 \\
& 3-1+2 P =0 \\
& P =-1 \\
& \left|\begin{array}{ccc}
\hat{ i } & \hat{ j } & \hat{ k } \\
1 & -1 & 2 \\
3 & 1 & -1
\end{array}\right|=-\hat{ i }+7 \hat{ j }+4 \hat{ k } \\
& \therefore(-\delta, 7 \delta, 4 \delta) \text { will lie on } L _3 \\
& \text { For } \delta=1 \text { the point will be }(-1,7,4)
\end{aligned}
\)

Hint

Let foot \(P (5 k -3,2 k +1,3 k -4)\)
DR’s \(\rightarrow A P: 5 k-4,2 k-1,3 k-7\)
DR’s \(\rightarrow\) Line \(: 5,2,3\)
Condition of perpendicular lines \((25 k-20)+(4 k-2)+(9 k-21)=0\)
Then \(k =\frac{43}{38}\)
Then \(19(\alpha+\beta+\gamma)=101\)

Hint

\(
\begin{aligned}
& \text { Area }=|\overrightarrow{ AC } \times \overrightarrow{ BD }| \\
& =\left|\begin{array}{ccc}
\hat{ i } & \hat{ j } & \hat{ k } \\
5 & -1 & 7 \\
1 & 2 & 3
\end{array}\right| \\
& =\frac{1}{2}|-17 \hat{ i }-8 \hat{ j }+11 \hat{ k }|=\frac{1}{2} \sqrt{474}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Direction ratio of } PR =(4,1,-1) \\
& \text { Direction ratio of } PQ =(1,4,-1) \\
& \text { Now, } \cos \theta=\left|\frac{4+4+1}{\sqrt{18} \cdot \sqrt{18}}\right| \\
& \theta=\frac{\pi}{3}
\end{aligned}
\)

Hint

\(
\text { Length of } O C=\frac{\sqrt{136}}{3}=\frac{2 \sqrt{34}}{3}
\)

Hint

Centroid \(G\) divides MR in \(1: 2\)
\(
G (1,2,2)
\)
Point of intersection \(A\) of given lines is \((2,-6,0)\)
\(
AG =\sqrt{69}
\)

Hint

\(
L _1=\frac{ x }{1}=\frac{ y -1}{2}=\frac{ z -2}{3}=\lambda
\)
\(
\begin{aligned}
& M (\lambda, 1+2 \lambda, 2+3 \lambda) \\
& \overrightarrow{P M}=(\lambda-1) \hat{i}+(1+2 \lambda) \hat{j}+(3 \lambda-5) \hat{k}
\end{aligned}
\)
\(\overrightarrow{ PM }\) is perpendicular to line \(L _1\)
\(
\begin{aligned}
& \overrightarrow{ PM } \cdot \overrightarrow{ b }=0 \quad(\overrightarrow{ b }=\hat{ i }+2 \hat{ j }+3 \hat{ k }) \\
& \Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0 \\
& 14 \lambda=14 \Rightarrow \lambda=1 \\
& \therefore M =(1,3,5) \\
& \overrightarrow{ Q }=2 \overrightarrow{ M }-\overrightarrow{ P } \quad[ M \text { is midpoint of } \overrightarrow{ P } \& \overrightarrow{ Q }] \\
& \overrightarrow{ Q }=2 \hat{ i }+6 \hat{ j }+10 \hat{ k }-\hat{ i }-7 \hat{ k } \\
& \overrightarrow{ Q }=\hat{ i }+6 \hat{ j }+3 \hat{ k } \\
& \therefore(\alpha, \beta, \gamma)=(1,6,3)
\end{aligned}
\)
Required line having direction cosine \((l, m, n)\)
\(
\begin{aligned}
& l^2+m^2+n^2=1 \\
& \Rightarrow l^2+\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2=1 \\
& l^2=\frac{1}{4}
\end{aligned}
\)
\(\therefore l=\frac{1}{2}\) [Line make acute angle with \(x\)-axis]
Equation of line passing through \((1,6,3)\) will be
\(
\overrightarrow{ r }=(\hat{ i }+6 \hat{ j }+3 \hat{ k })+\mu\left(\frac{1}{2} \hat{ i }-\frac{1}{2} \hat{ j }-\frac{1}{\sqrt{2}} \hat{ k }\right)
\)
Option (3) satisfying for \(\mu=4\)

Hint

\(
B =(2 \lambda+7,-3 \lambda-2,6 \lambda+11)
\)
Point \(B\) lies on \(\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}\)
\(
\begin{aligned}
& \frac{2 \lambda+7-6}{1}=\frac{-3 \lambda-2-4}{0}=\frac{6 \lambda+11-8}{3} \\
& -3 \lambda-6=0 \\
& \lambda=-2 \\
& B \Rightarrow(3,4,-1) \\
& AB =\sqrt{(7-3)^2+(4+2)^2+(11+1)^2} \\
& =\sqrt{16+36+144} \\
& =\sqrt{196} \\
& =14
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \\
& \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}
\end{aligned}
\)
the shortest distance between the lines
\(
\begin{aligned}
& =\left|\frac{\overrightarrow{ a }-\overrightarrow{ b }) \cdot\left(\overrightarrow{ d _1} \times \overrightarrow{ d _2}\right)}{\left|\overrightarrow{ d _1} \times \overrightarrow{ d _2}\right|}\right| \\
& =\left|\frac{\left|\begin{array}{ccc}
\lambda-4 & 0 & 2 \\
1 & 2 & -3 \\
2 & 4 & -5
\end{array}\right|}{\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -3 \\
2 & 4 & -5
\end{array}\right|}\right|
\end{aligned}
\)
\(
\begin{aligned}
& =\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{|2 \hat{i}-1 \hat{j}+0 \hat{k}|}\right| \\
& \frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right| \\
& 3=|\lambda-4| \\
& \lambda-4= \pm 3 \\
& \lambda=7,1
\end{aligned}
\)
Sum of all possible values of \(\lambda\) is \(=8\)

Hint

The equation of the plane passing through points \(A(-1,-2,-3), B(9,3,4)\), and \(C(9,-2,1)\) can be written using the determinant :
\(
\left|\begin{array}{ccc}
x+1 & y+2 & z+3 \\
10 & 5 & 7 \\
10 & 0 & 4
\end{array}\right|=0
\)
Expanding the determinant, we get :
\(
2 x+3 y-5 z-7=0
\)
Next, we find the foot of the perpendicular from point \(P(3,-2,-9)\) to the plane. Using the coordinates of \(P\) and the equation of the plane, we can find the ratio of the perpendicular distance to the sum of the squares of the coefficients of \(x, y\), and \(z\) :
\(
\frac{2(3)+3(-2)-5(-9)-7}{2^2+3^2+(-5)^2}=\frac{-38}{38}
\)
We can now find the coordinates of the foot of the perpendicular, \(Q(\alpha, \beta, \gamma)\) :
\(
\frac{\alpha-3}{2}=\frac{\beta+2}{3}=\frac{\gamma+9}{-5}=-\frac{38}{38}
\)
Solving for \(\alpha, \beta\), and \(\gamma\) :
\(
\begin{aligned}
& \alpha=3-2\left(-\frac{38}{38}\right)=1 \\
& \beta=-2+3\left(-\frac{38}{38}\right)=-5 \\
& \gamma=-9-5\left(-\frac{38}{38}\right)=-4
\end{aligned}
\)
So, the coordinates of the foot of the perpendicular are \(Q(1,-5,-4)\). Now, we can find the distance of point \(Q\) from the origin :
\(
O Q=\sqrt{\alpha^2+\beta^2+\gamma^2}=\sqrt{1^2+(-5)^2+(-4)^2}=\sqrt{42}
\)

Hint

\(
\begin{aligned}
& -x+2 y-9 z=7-(1) \\
& -x+3 y-7 z=9-(2) \\
& -2 x+y+5 z=8-(3) \\
& (2)-(1) \\
& y+16 z=2 \dots(4)\\
& (3)-2 \times(1) \\
& -3 y+23 z=-6-(5) \\
& 3 \times(4)+(5) \\
& 71 z=0 \Rightarrow z=0 \\
& y=2 \\
& (-3,2,0) \rightarrow(\alpha, \beta, \gamma) \\
& \text { Put in }-3 x+y+13 z=\lambda \\
& \lambda=9+2=11
\end{aligned}
\)
The equation of the plane is:
\(
2 x-2 y+z=\lambda
\)
We already know that \(\lambda=11\). So the equation of the plane becomes:
\(
2 x-2 y+z=11
\)
Now, let’s find the distance \(d\) of the point \((\alpha, \beta, \gamma)=(-3,2,0)\) from the plane using the formula :
\(
d=\frac{|A x+B y+C z+D|}{\sqrt{A^2+B^2+C^2}}
\)
Plugging in the values for the point and the plane equation, we get:
\(
d=\frac{|2(-3)-2(2)+0-11|}{\sqrt{2^2+(-2)^2+1^2}}=\frac{|-6-4-11|}{\sqrt{4+4+1}}=\frac{21}{\sqrt{9}}=7
\)
So, the distance of the point \((-3,2,0)\) from the plane \(2 x-2 y+z=11\) is 7.

Hint

Given the two lines :
\(
\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1} \frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}
\)
Let’s find the direction vectors of these lines: \(\vec{d}_1=\langle 0,4,1\rangle\) and \(\vec{d}_2=\langle 3,-4,0\rangle\).
Now, let’s find the cross product of the direction vectors, which will give a vector that is perpendicular to both lines :
\(
\vec{n}=\vec{d}_1 \times \vec{d}_2=\langle 4,3,-12\rangle
\)
Let’s find the vector connecting a point on line 1 to a point on line 2 :
\(
\vec{c}=\langle 2 \lambda, 3,-12\rangle
\)
The shortest distance between the two lines is the projection of \(\vec{c}\) onto \(\vec{n}\) :
\(
d=\left|\frac{\vec{c} \cdot \vec{n}}{|\vec{n}|}\right|=\left|\frac{(2 \lambda)(4)+(3)(3)-(12)(-12)}{\sqrt{16+9+144}}\right|
\)
We are given that the shortest distance is 13 :
\(
\begin{aligned}
& 13=\left|\frac{8 \lambda+153}{13}\right| \\
& |8 \lambda+153|=169
\end{aligned}
\)
We have two cases:
1. \(8 \lambda+153=169\)
\(
\lambda=\frac{16}{8}
\)
2. \(8 \lambda+153=-169[latex]
[latex]
\lambda=\frac{-322}{8}
\)
Now, let’s calculate \(8\left|\sum_{\lambda \in S} \lambda\right|\) :
\(
8\left|\frac{16}{8}+\frac{-322}{8}\right|=8\left|\frac{-306}{8}\right|=306
\)

Hint

Given two lines:
\(
\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}
\)
and \(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\)
These lines are coplanar if the determinant of the matrix
\(
\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0
\)
Now let’s apply this condition to the given problem. The given line is :
\(
\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}
\)
So, the coordinates of any point on this line are \((-3,1,5)\) and the direction ratios are \((-3,1,5)\).
Now, let’s calculate the determinants for each option and check which one equals zero.
For Option A :
\(
\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{4}
\)
The coordinates of any point on this line are \((-1,2,5)\) and the direction ratios are \((-1,2,4)\).
The determinant is :
\(
\left|\begin{array}{ccc}
-1-(-3) & 2-1 & 5-5 \\
-3 & 1 & 5 \\
-1 & 2 & 4
\end{array}\right|=\left|\begin{array}{ccc}
2 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 4
\end{array}\right|
\)
Applying the formula :
\(
\begin{aligned}
& =2(1 \times 4-5 \times 2)-1((-3 \times 4)-(5 \times-1))+0((-3 \times 2)-(1 \times-1)) \\
& =2(4-10)-1(-12-(-5))+0(-6-(-1)) \\
& =2(-6)-1(-7)+0(-5) \\
& =-12+7+0 \\
& =-5
\end{aligned}
\)
The determinant for Option A is not equal to zero, so this line is not coplanar with the given line.
For Option B, we have :
\(
\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}
\)
The coordinates of any point on this line are \((-1,2,5)\) and the direction ratios are \((-1,2,5)\).
\(
\left|\begin{array}{ccc}
-1-(-3) & 2-1 & 5-5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{array}\right|=\left|\begin{array}{ccc}
2 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{array}\right|
\)
Applying the formula, we get:
\(
=2(5-10)-1(-15-(-5))+0(-6-(-1))=0
\)
So, the determinant for Option B equals zero, which confirms that the line in Option B is coplanar with the given line.
For Option C :
\(
\frac{x-1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}
\)
The coordinates of any point on this line are \((1,2,5)\) and the direction ratios are \((-1,2,5)\).
The determinant is:
\(
\left|\begin{array}{ccc}
1-(-3) & 2-1 & 5-5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{array}\right|=\left|\begin{array}{ccc}
4 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{array}\right|
\)
Applying the formula :
\(
=4(5-10)-1(-15-(-5))+0(-6-(-1))=-10
\)
For Option D:
\(
\frac{x+1}{1}=\frac{y-2}{2}=\frac{z-5}{5}
\)
The coordinates of any point on this line are \((-1,2,5)\) and the direction ratios are \((1,2,5)\).
The determinant is :
\(
\left|\begin{array}{ccc}
-1-(-3) & 2-1 & 5-5 \\
-3 & 1 & 5 \\
1 & 2 & 5
\end{array}\right|=\left|\begin{array}{ccc}
2 & 1 & 0 \\
-3 & 1 & 5 \\
1 & 2 & 5
\end{array}\right|
\)
Applying the formula :
\(
=2(5-10)-1(-15-5)+0(-6-1)=10
\)
So the determinant for Option \(D\) is not equal to zero, which means the line in Option D is not coplanar with the given line.

Hint

The first step is to find the normal vector to the desired plane. Since the plane is parallel to the line passing through the points \((5,1,-7)\) and \((1,-1,-1)\), the direction vector of that line is also parallel to the plane. The direction vector is the difference between the coordinates of the two points, which is \((5-1,1-(-1),-7-(-1))=(4,2,-6)\).
Next, let’s find another vector that is parallel to the plane. This vector can be obtained by taking the difference between the coordinates of the points \((0,-1,2)\) and \((-1,2,1)\), which the plane passes through. This gives us a vector of \((0-(-1),-1-2,2-1)=(1,-3,1)\).
The normal to the plane is perpendicular to both these vectors. It can be found by taking the cross product of the two vectors.
The cross product of vectors \((4,2,-6)\) and \((1,-3,1)\) is :
\(
\begin{aligned}
\vec{n} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
4 & 2 & -6 \\
1 & -3 & 1
\end{array}\right|=\hat{i}(-16)-\hat{j}(+10)+\hat{k}(-14) \\
& =-16 \hat{i}-10 \hat{j}-14 \hat{k}
\end{aligned}
\)
The equation of the plane can now be written in the form :
\(
-16 x-10 y-14 z=d
\)
We can find the constant ‘d’ by substituting one of the points through which the plane passes, say \((0,-1,2)\) :
\(
d=-16 \cdot 0-10 .(-1)-14 \cdot 2=10-28=-18
\)
So, the equation of the plane is \(-16 x-10 y-14 z=-18\).
Now, we substitute the given options into the equation to check which one satisfies it :
Option A: \((-16.0-10.5-14 .(-2)=-18)=>-50+28=-22 \neq-18\), so A is not correct.
Option B: \((-16.2-10.0-14.1=-18)=>-32-14=-46 \neq-18\), so B is not correct.
Option C: \((-16.1-10 .(-2)-14.1=-18)=>-16+20-14=-10 \neq-18\), so \(C\) is not correct.
Option D: \((-16 .(-2)-10.5-14.0=-18)=>32-50=-18\), so \(D\) is correct.
So, the plane also passes through the point given in Option D, which is \((-2,5,0)\).

Hint

Equation of line \(\frac{x-4}{4-1}=\frac{y-5}{5-(-7)}=\frac{z-8}{8-5}\)
\(
\frac{x-4}{3}=\frac{y-5}{12}=\frac{z-8}{3}
\)
Let point \(N (3 \lambda+4,12 \lambda+5,3 \lambda+8)\)
\(
\begin{aligned}
& \overrightarrow{ PN }=(3 \lambda+4-1) \hat{i}+(12 \lambda+5-(-2)) \hat{j}+(3 \lambda+8-3) \hat{ k } \\
& \overrightarrow{ PN }=(3 \lambda+3) \hat{i}+(12 \lambda+7) \hat{ j }+(3 \lambda+5) \hat{ k }
\end{aligned}
\)
And parallel vector to line (say \(\vec{a}=3 \hat{i}+12 \hat{j}+3 \hat{k}\) )
Now, \(\overrightarrow{ PN } \cdot \overrightarrow{ a }=0\)
\(
\begin{aligned}
& (3 \lambda+3) 3+(12 \lambda+7) 12+(3 \lambda+5) 3=0 \\
& 162 \lambda+108=0 \Rightarrow \lambda=\frac{-108}{162}=\frac{-2}{3}
\end{aligned}
\)
So point \(N\) is \((2,-3,6)\)
Now distance of \(N (2,-3,6)\) from \(2 x-2 y+z+5=0\) is
\(
=\left|\frac{2(2)-2(-3)+6+5}{\sqrt{4+4+1}}\right|=7
\)

Hint

Given the equation of the plane passing through the intersection of the two given planes:
\(
\begin{aligned}
& P:(x+2 y+a z-2)+\lambda(x-y+z-3)=0 \\
& \Rightarrow x(\lambda+1)+y(2-\lambda)+z(a+\lambda)-2-3 \lambda=0
\end{aligned}
\)
This is the same as the given equation \(5 x-11 y+b z=6 a-1\).
Now, comparing the coefficients of the corresponding variables in both equations:
\(
\frac{\lambda+1}{5}=\frac{2-\lambda}{-11}=\frac{a+\lambda}{b}=\frac{2+3 \lambda}{6 a-1}
\)
Solving for \(\lambda\) :
\(
\begin{aligned}
& -11 \lambda-11=10-5 \lambda \\
& 6 \lambda=-21 \Rightarrow \lambda=-\frac{7}{2}
\end{aligned}
\)
Now, substituting the value of \(\lambda\) back into the equations:
\(
\frac{2-\lambda}{-11}=\frac{2+3 \lambda}{6 a-1} \Rightarrow \frac{2+\frac{7}{2}}{-11}=\frac{2-\frac{21}{2}}{6 a-1}
\)
From this equation, we find the value of a :
\(
6 a-1=17 \Rightarrow a=3
\)
Now, substituting the value of \(a\) and \(\lambda\) into the equation:
\(
\begin{aligned}
& \frac{2-\lambda}{-11}=\frac{a+\lambda}{b} \Rightarrow-\frac{1}{2}=\frac{3-\frac{7}{2}}{b} \\
& \Rightarrow-\frac{b}{2}=-\frac{1}{2} \Rightarrow b=1
\end{aligned}
\)
Therefore, the point \((a,-c, c) \equiv(3,-c, c)\).
The given distance is \(\frac{2}{\sqrt{a}}=\frac{2}{\sqrt{3}}\).
The plane is: \(5 x-11 y+z=17\).
Now, let’s find the distance:
\(
\begin{aligned}
& \left|\frac{15+11 c+c-17}{\sqrt{147}}\right|=\frac{2}{\sqrt{3}} \\
& \Rightarrow c=-1, \frac{4}{3}
\end{aligned}
\)
Since \(c \in Z\), we have \(c=-1\).
Therefore, \(\frac{a+b}{c}=\frac{3+1}{-1}=-4\).

Hint

Step 1: Determine the line of shortest distance between the given two lines:
Direction vector of line \(1: \vec{d}_1=2 \hat{i}+\hat{k}\)
Direction vector of line 2 : \(\overrightarrow{d_2}=\hat{i}-\hat{j}+\hat{k}\)
Now, let’s find the cross product \(\vec{N}=\vec{d}_1 \times \vec{d}_2\)
\(
\vec{N}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 0 & 1 \\
1 & -1 & 1
\end{array}\right|=(\hat{i}(0+1)-\hat{j}(2-1)+\hat{k}(-2-0))=\hat{i}-\hat{j}-2 \hat{k}
\)
So, the direction vector of the line of shortest distance is \(\vec{N}=\hat{i}-\hat{j}-2 \hat{k}\).

Step 2: Find the equation of a line passing through point \(A(-1,2,3)\) and having the direction vector \(\vec{N}\) :
\(
\frac{x+1}{1}=\frac{y-2}{-1}=\frac{z-3}{-2}=\lambda
\)

Step 3: Find the point \(P\) where the line intersects the given plane:
Let the coordinates of point \(P\) be in terms of \(\lambda\) :
\(
P(\lambda-1,-\lambda+2,-2 \lambda+3)
\)
Since \(P\) lies on the plane \(x-2 y+3 z=10\), we can substitute the coordinates of \(P\) in terms of \(\lambda\) into the equation of the plane:
\(
\begin{aligned}
& (\lambda-1)-2(-\lambda+2)+3(-2 \lambda+3)=10 \\
& \lambda=-2 \\
& P(-3,4,7)
\end{aligned}
\)
Step 4: Calculate the distance between points \(A\) and \(P\) :
\(
A P=\sqrt{(-3-(-1))^2+(4-2)^2+(7-3)^2}=2 \sqrt{6}
\)

Hint

\(
\begin{aligned}
& (3 x +2 y + z -2)+\mu( x -3 y +2 z -13)=0 \\
& 3(3+\mu)+1 \cdot(2-3 \mu)-2(1+2 \mu)=0 \\
& 9-4 \mu=0 \\
& \mu=\frac{9}{4} \\
& 4(-15-8+\alpha-2)+9(-5+12+2 \alpha-13)=0 \\
& -100+4 \alpha-54+18 \alpha=0 \\
& \Rightarrow \alpha=7 \\
& \text { Let } P \equiv(3 \lambda-5, \lambda-4,-2 \lambda+7) \\
& \text { Direction ratio of } PQ (3 \lambda-1, \lambda-1,-2 \lambda+5) \\
& \text { But PQ } \perp \ell_1 \\
& \Rightarrow 3(3 \lambda-1)+1 \cdot(\lambda-1)-2(-2 \lambda+5)=0 \\
& \Rightarrow \lambda=1 \\
& P (-2,-3,5) \Rightarrow| a |+| b |+| c |=10
\end{aligned}
\)

Hint

Equation of plane after rotation :
\(
\begin{aligned}
& (4 x-y+z-10)+\lambda(x+y-z-y)=0 \\
\Rightarrow & (4+\lambda) x+y(\lambda-1)+z(1-\lambda)-4 \lambda-10=0 \\
& \overrightarrow{n_1} \cdot \overrightarrow{n_2}=0 \\
\Rightarrow & (4+\lambda) 4+(\lambda-1)(-1)+(1-\lambda) 1=0 \\
\Rightarrow & 16+4 \lambda-\lambda+1+1-\lambda=0 \\
\Rightarrow & 2 \lambda=-18 \\
\Rightarrow & \lambda=-9
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \text { equation of plane : }-5 x-10 y+10 z+26=0 \\
& \text { Distance of plane from }(2,3,-4) \\
& \qquad=\left|\frac{-10-30-40+26}{\sqrt{100+100+26}}\right|=\frac{54}{15}=\alpha \\
& \therefore 35 \alpha=35 \cdot \frac{54}{15}=7 \times \frac{54}{3}=7 \times 18=126
\end{aligned}
\)

Hint

Equation of line \(P Q\) :
\(
\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda
\)
Let \(R\) be \((3 \lambda+2,4 \lambda-1,2 \lambda+2)\)
\(R\) lies on plane \(x-y+z=4\)
\(
\begin{aligned}
& \therefore \quad 3 \lambda+2-4 \lambda+1+2 \lambda+2=4 \\
& \Rightarrow \quad \lambda=-1 \\
& \therefore \quad R(-1,-5,0)
\end{aligned}
\)
Let \(S R\) be \(: \frac{x+1}{2}=\frac{y+5}{2}=\frac{z-0}{1}=k\)
\(
\begin{aligned}
& \text { Point } S:(2 k-1,2 k-5, k) \\
& S \text { lies on plane: } x+2 y+3 z+2=0 \\
& \Rightarrow(2 k-1)+(4 k-10)+3 k+2=0 \\
& \Rightarrow 9 k-9=0 \Rightarrow k=1 \\
& \therefore S=(1,-3,1) \\
& \therefore S R=\sqrt{4+4+1}=3
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \overrightarrow{A B}=8 \hat{i}-2 \hat{k} \\
& \overrightarrow{A C}=-4 \hat{i}+3 \hat{j}+2 \hat{k} \\
& \overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
8 & 0 & -2 \\
-4 & 3 & 2
\end{array}\right| \\
& =6 \hat{i}-8 \hat{j}+24 \hat{k}
\end{aligned}
\)
Equation of plane : \(6 x-8 y+24 z=d\) passes through \((5,3,0)\)
\(
\begin{aligned}
& 6 \times 5-8 \times 3+24 \times 0=d \\
& \Rightarrow d=6
\end{aligned}
\)
\(
\begin{aligned}
& 6 x-8 y+24 z=6 \\
& \Rightarrow 3 x-4 y+12 z=3
\end{aligned}
\)
Distance of point \((3,4, \alpha)\)
\(
\frac{9-16+12 \alpha-3}{\sqrt{9+16+144}}=2 \Rightarrow \alpha=3
\)
Distance of point \((2, \alpha, a)\)
\(
\begin{aligned}
& \frac{3 \times 2-4 \times 3+12 \times a-3}{13}=3 \\
& \Rightarrow 12 a-9=39 \\
& \Rightarrow 12 a=48 \\
& \Rightarrow a=4
\end{aligned}
\)

Hint

Co-ordinate of image of point \(P(2,3,5)\) in the plane \(2 x+y-3 z=6\) is
\(
\begin{aligned}
\frac{\alpha-2}{2} & =\frac{\beta-3}{1}=\frac{\gamma-5}{-3}=\frac{-2(2 \times 2+3-3 \times 5-6)}{2^2+1^2+(-3)^2} \\
\frac{\alpha-2}{2} & =\frac{\beta-3}{1}=\frac{\gamma-5}{-3}=2 \\
\frac{\alpha-2}{2} & =2, \frac{\beta-3}{1}=2, \frac{\gamma-5}{-3}=2 \\
\alpha & =6, \beta=5, \gamma=-1
\end{aligned}
\)
Hence, \(\alpha+\beta+\gamma=6+5-1=10\)

Hint

The equation of plane that passes through the point \((-2,3,5)\) is
\(
a(x+2)+b(y-3)+c(z-5)=0 \dots(i)
\)
The plane is perpendicular to
\(
\begin{aligned}
& 2 x+4 y+5 z=8 \text { and } 3 x-2 y+3 z=5 \\
& \therefore 2 a+4 b+5 c=0 \ldots \ldots \ldots(i i)
\end{aligned}
\)
and \(3 a-2 b+3 c=0\). (iii)
\(
\begin{aligned}
& \therefore \frac{ a }{\left|\begin{array}{cr}
4 & 5 \\
-2 & 3
\end{array}\right|}=\frac{- b }{\left|\begin{array}{ll}
2 & 5 \\
3 & 3
\end{array}\right|}=\frac{ c }{\left|\begin{array}{cc}
2 & 4 \\
3 & -2
\end{array}\right|} \\
& \Rightarrow \frac{ a }{22}=\frac{ b }{9}=\frac{ c }{-16}
\end{aligned}
\)
\(\therefore\) From Equation (i) equation of plane
\(
\begin{aligned}
& 22(x+2)+9(y-3)-16(z-5)=0 \\
& \Rightarrow 22 x+9 y-16 z+97=0
\end{aligned}
\)
Here, \(\alpha=22, \beta=9, \gamma=-16\)
\(
\therefore \alpha+\beta+\gamma=22+9-16=15
\)

Hint

Equation of plane passing through the points \(A(1,2,0), B(1,4,1)\) and \(C(0,5,1)\) is
\(
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-2 & z-0 \\
0 & 2 & 1 \\
-1 & 3 & 1
\end{array}\right|=0 \\
& \Rightarrow x+y-2 z=3
\end{aligned}
\)
Now \(Q(\alpha, \beta, \gamma)\) is the image of the point \(P(1,2,6)\) in the plane \(x+y-2 z-3=0\)
\(
\begin{aligned}
& \therefore \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=\frac{-2[1+2-2(6)-3]}{1^2+1^2+(-2)^2} \\
& \Rightarrow \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=4 \\
& \Rightarrow \alpha=5, \beta=6, \gamma=-2
\end{aligned}
\)
Hence, \(\alpha^2+\beta^2+\gamma^2=5^2+6^2+(-2)^2=65\)

Hint

We have, \(\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}\) intersect the line \(\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}\) and \(\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}\)
Now, \(\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}=\lambda \dots(i)\).
\(
\Rightarrow x=\lambda, y=6-2 \lambda, z=5 \lambda-8
\)
Also, \(\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}=k \dots(ii)\).
\(\Rightarrow x=4 k+5, y=3 k+7, z=k-2\)
\(\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}=\mu \ldots \ldots \ldots \ldots(i i i)\)
\(
\Rightarrow x=6 \mu-3, y=3-3 \mu, z=\mu+6
\)
On solving Eqs. (i) and (ii), we get \(\lambda=1, k=-1\)
\(\therefore[latex] Co-ordinate of [latex]A\) is \((1,4,-3)\)
On solving Eqs. (i) and (iii), we get \(\lambda=3, \mu=1\)
\(\therefore\) Co-ordinate of \(\beta\) is \((3,0,7)\)
Co-ordinate of mid-point of \(A B\) is \(\left(\frac{1+3}{2}, \frac{4+0}{2}, \frac{-3+7}{2}\right)\) or \((2,2,2)\)
Perpendicular distance of mid-point of \(A B\) from the plane \(2 x-2 y+z=14\) is
\(
\frac{|2(2)-2(2)+2-14|}{\sqrt{2^2+(-2)^2+1^2}}=4
\)

Hint

Given, the lines are
\(
\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2} \dots(i)
\)
and \(\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0} \dots(ii)\)
Formula for shortest distance between two skew-lines,
\(
\begin{aligned}
& S D=\left|\frac{\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|}{\left|\begin{array}{ccc}
\hat{ i } & \hat{ j } & \hat{ k } \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|}\right|=\left|\frac{\left|\begin{array}{ccc}
6 & 1 & -8 \\
1 & -2 & 2 \\
1 & 2 & 0
\end{array}\right|}{\left|\begin{array}{ccc}
\hat{ i } & \hat{ j } & \hat{ k } \\
1 & -2 & 2 \\
1 & 2 & 0
\end{array}\right|}\right| \\
& =\left|\frac{6(-4)-1(-2)-8(4)}{|-4 \hat{ i }+2 \hat{ j }+4 \hat{ k }|}\right| \\
& =\left|\frac{-24+2-32}{\sqrt{36}}\right| \\
& =\left|\frac{-54}{6}\right|=|-9|=9 \\
&
\end{aligned}
\)

Hint

Given that two vertex of a \(\triangle A B C\) be \(A(2,4,6)\) and \(B(0,-2,-5)\) and \(G=\) centroid \(=(2,1,-1)\) [Given]
Let, the other vertex is \(C(x, y, z)\)
According to the question,
\(
\begin{aligned}
& \frac{2+0+x}{3}=2 \\
& \Rightarrow x=4 \\
& \frac{4-2+y}{3}=1 \\
& \Rightarrow y=1 \\
& \frac{6-5+z}{3}=-1 \\
& \Rightarrow z=1
\end{aligned}
\)
Hence, third vertex is \(C(4,1,-4)\)
Now, if image of \(C(4,1,-4)\) in the plane \(x+2 y+4 z=11\) is \(D(\alpha, \beta, \gamma)\).
So, \(\frac{\alpha-4}{1}=\frac{\beta-1}{2}=\frac{\gamma+4}{4}=k\) (say)

\(
\Rightarrow \alpha=k+4, \beta=2 k+1, \gamma=4 k-4
\)
Then, co-ordinates of \(D\) is \((k+4,2 k+1,4 k-4)\)
Let, \(E\) be the mid-point of \(C D\), which lies on the plane \(x+2 y+4 z=11\).
Then, co-ordinates of \(E\) is
\(
\left(\frac{k+8}{2}, \frac{2 k+2}{2}, \frac{4 k-8}{2}\right)=\left(\frac{k+8}{2}, k+1,2 k-4\right)
\)
Since, \(E\) lies on the plane, \(x+2 y+4 z-11=0\)
\(
\begin{aligned}
& \text { So, } \frac{k+8}{2}+2 k+2+8 k-16-11=0 \\
& \Rightarrow \frac{k+8+4 k+4+16 k-54}{2}=0 \\
& \Rightarrow 21 k=42 \\
& \Rightarrow k=2
\end{aligned}
\)
Hence, \(D \equiv(6,5,4) \equiv(\alpha, \beta, \gamma)\)

So, \(\alpha=6, \beta=5, \gamma=4\)
Now, \(\alpha \beta+\beta \gamma+\gamma \alpha=6 \times 5+5 \times 4+4 \times 6\)
\(
=30+20+24=74
\)

Hint

Given, equation of line is
\(
\begin{aligned}
& \frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}=k \\
& \therefore x=3 k-3, y=k-2, z=1-2 k
\end{aligned}
\)
Since, given that \(P \equiv(3 k-3, k-2,1-2 k)\) be the point of intersection of the given line and the plane \(x+y+z=2\)
So, \((3 k-3)+(k-2)+(1-2 k)=2\)
\(
\Rightarrow 2 k-4=2 \Rightarrow k=3
\)
Thus, \(P=(6,1,-5)\)
Now, distance of point \(P\) from the plane \(3 x-4 y+12 z=32\) is
\(
\begin{aligned}
& q=\left|\frac{18-4-60-32}{\sqrt{9+16+144}}\right|=\left|\frac{-78}{\sqrt{169}}\right| \\
& \Rightarrow q=\left|\frac{-78}{13}\right|=\frac{78}{13}=6 \\
& \therefore q=6 \Rightarrow 2 q=12
\end{aligned}
\)
Thus, \(q\) and \(2 q\) are roots of the equation \(x^2-18 x+72=0\)

Hint

We have, \(\theta=\cos ^{-1} \frac{1}{3}\)
\(
\begin{aligned}
& \Rightarrow \cos \theta=\frac{1}{3} \\
& \therefore \sin \theta=\sqrt{1-\left(\frac{1}{3}\right)^2}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}
\end{aligned}
\)
The given plane line and are
\(
\begin{aligned}
& a x+y-z=b \\
& x-1=a-y=z+1 \\
& \therefore \sin \theta=\frac{a \cdot 1+(1)(-1)+(-1)(1)}{\sqrt{a^2+1^2+1^2} \sqrt{1^2+1^2+1^2}} \\
& \Rightarrow \frac{a-1-1}{\sqrt{a^2+2} \sqrt{3}}=\frac{2 \sqrt{2}}{3} \\
& \Rightarrow 3(a-2)=2 \sqrt{6} \sqrt{a^2+2} \\
& \Rightarrow 9\left(a^2+4-4 a\right)=24\left(a^2+2\right) \\
& \Rightarrow 9 a^2+36-36 a=24 a^2+48 \\
& \Rightarrow 15 a^2+36 a+12=0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow 5 a^2+12 a+4=0 \\
& \Rightarrow 5 a^2+10 a+2 a+4=0 \\
& \Rightarrow 5 a(a+2)+2(a+2)=0 \\
& \Rightarrow a=\frac{-2}{5},-2
\end{aligned}
\)
So, \(a=-2[\because a \in Z]\)
Hence, the eqn. of plane is \(-2 x+y-z-b=0\)
\(
\begin{aligned}
& \text { Now, } d=\left|\frac{-12-6-4-b}{\sqrt{4+1+1}}\right|=3 \sqrt{6} \\
& \Rightarrow|-(b+22)|=18 \\
& \Rightarrow b=18-22=-4 \\
& \therefore a^4+b^2=(-2)^4+(-4)^2 \\
& =16+16=32
\end{aligned}
\)

Hint

Equation of line : \(\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}\)
Let \(B \equiv(2,4,-3)\)
\(
\begin{aligned}
& \text { So, } \overrightarrow{ AB }=(2-1) \hat{i}+(4-2) \hat{j}+(-3+5) \hat{k} \\
& =\hat{i}+2 \hat{j}+2 \hat{k} \\
& \vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -3 & 7 \\
1 & 2 & 2
\end{array}\right|=(-6-14) \hat{i}-(2-7) \hat{j}+(2+3) \hat{k} \\
& =-20 \hat{i}+5 \hat{j}+5 \hat{k} \\
& =-5(4 \hat{i}-\hat{j}-\hat{k})
\end{aligned}
\)
\(\therefore\) Equation of plane is :
\(
\begin{aligned}
& 4(x-1)+(-1)(y-2)-1(z+5)=0 \\
& \Rightarrow 4 x-4-y+2-z-5=0 \\
& \Rightarrow 4 x-y-z-7=0
\end{aligned}
\)
\(\because\) Image of point \((-1,3,4)\) is \((\alpha, \beta, \gamma)\)
So, \(\frac{\alpha+1}{4}=\frac{\beta-3}{-1}=\frac{\gamma-4}{-1}=\frac{-2(-4-3-4-7)}{16+1+1}=2\)
\(
\Rightarrow \alpha=7, \beta=1, \gamma=2
\)
So, \(\alpha+\beta+\gamma=10\)

Hint

The given lines are
\(\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}\) and \(\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}\)
So, \(\vec{b}_1=4 \hat{i}+5 \hat{j}+3 \hat{k}\)
\(\vec{b}_2=3 \hat{i}+4 \hat{j}+2 \hat{k}\)
\(\vec{a}_1=4 \hat{i}-2 \hat{j}-3 \hat{k}\)
\(\vec{a}_2=\hat{i}+3 \hat{j}+4 \hat{k}\)
\(\therefore \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2\end{array}\right|\)
\(=(10-12) \hat{i}-(8-9) \hat{j}+(16-15) \hat{k}\)
\(=-2 \hat{i}+\hat{j}+\hat{k}\)
\(
\begin{aligned}
& \text { Shortest distance, } d=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right| \\
& =\left|\frac{(3 \hat{i}-5 \hat{j}-7 \hat{k}) \cdot(-2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{4+1+1}}\right| \\
& =\left|\frac{-6-5-7}{\sqrt{6}}\right|=\frac{18}{\sqrt{6}}=3 \sqrt{6} \text { units }
\end{aligned}
\)

Hint

Equation of plane \(P\) containing the given lines is
\(
\begin{aligned}
& (x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0 \\
& \Rightarrow(1+2 \lambda) x+(2+\lambda) y+(3-\lambda) z+(-4+5 \lambda)=0
\end{aligned}
\)
Now, plane \(P\) is perpendicular to plane \(P ^{\prime}\)
\(
\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})
\)
So, normal to plane \(P ^{\prime}\) is
\(
\begin{aligned}
& \vec{n}=(\hat{i}+\hat{j}+\hat{k}) \times(\hat{i}-2 \hat{j}+3 \hat{k}) \\
& \Rightarrow \vec{n}=5 \hat{i}-2 \hat{j}-3 \hat{k}
\end{aligned}
\)
\(\therefore P\) and \(P ^{\prime}\) are perpendicular
\(
\begin{aligned}
& \therefore 5(1+2 \lambda)-2(2+\lambda)-3(3-\lambda)=0 \\
& \Rightarrow 5+10 \lambda-4-2 \lambda-9+3 \lambda=0 \\
& \Rightarrow 11 \lambda=8 \Rightarrow \lambda=\frac{8}{11}
\end{aligned}
\)
\(
\therefore P:\left(1+\frac{16}{11}\right) x+\left(2+\frac{8}{11}\right) y+\left(3-\frac{8}{11}\right) z+\left(5 \times \frac{8}{11}-4\right)=0
\)
i.e., \(27 x+30 y+25 z=4\)
which is same as \(a x+b y+c z=4\)
\(\therefore a=27, b=30\) and \(c=25\)
\(
\Rightarrow a-b+c=27-30+25=22
\)

Hint

Given plane \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6\) and
\(
\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5
\)
Equation of plane passing through both plane
\(
\begin{aligned}
& P _1 \rightarrow(x \hat{i}+y \hat{j}+2 \hat{k})(\hat{i}+\hat{j}+\hat{k})=6 \\
& P _1=x+y+z=6 \\
& P _2 \rightarrow(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5 \\
& P _2 \rightarrow=2 x+3 y+4 z=-5 \\
& P _1+\lambda P _2=0 \\
& \Rightarrow(x+y+z-6)+\lambda(2 x+3 y+4 z+5)=0
\end{aligned}
\)
passes through \((0,2,-2)\)
\(
\begin{aligned}
& \Rightarrow(0+2-2-6)+\lambda(2 \times 0+3 \times 2+4 \times(-2)+5)=0 \\
& \Rightarrow \lambda=2
\end{aligned}
\)
Equation of plane
\(
5 x+7 y+9 z+4=0
\)
\(\therefore\) Distance of the point \((12,12,18)\) from the plane
\(
\begin{aligned}
d & =\frac{5 \times 12+7 \times 12+9 \times 18+4}{\sqrt{5^2+7^2+9^2}} \\
& =\frac{60+84+162+4}{\sqrt{155}}=\frac{310}{\sqrt{155}} \\
\therefore d^2 & =\frac{310 \times 310}{155}=620
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \Rightarrow[(1+2 \lambda) \hat{ i }+(1+3 \lambda) \hat{ j }+(1+4 \lambda) \hat{ k }] \\
& \quad(2 \hat{ i }+\hat{ j }-3 \hat{ k }) \\
& 2+4 \lambda+1+3 \lambda-3-12 \lambda=0 \\
& 5 \lambda=0 \Rightarrow \lambda=0
\end{aligned}
\)
Line \(\overrightarrow{ AB }, \overrightarrow{ r }=\hat{ j }+2 \hat{ k }+\mu(\hat{ i }+\hat{ j }+\hat{ k })\)
General form: \(Q (\mu, 1+\mu, 2+\mu)\)
\(
\begin{aligned}
& \therefore \overrightarrow{ PQ } \cdot \overrightarrow{ AB }=0 \\
& (\mu-1)+(10+\mu)+\mu=0 \\
& 3 \mu=-9 \Rightarrow \mu=-3
\end{aligned}
\)
\(
\therefore \text { distance }=\sqrt{16+49+9}=\sqrt{74}
\)

Hint

Equation of plane intersection of two plane
\(
\begin{aligned}
& P _1+\lambda P _2=0 \\
& P _1: 2 x-y+z=3 \text {, and } P _2: 4 x-3 y+5 z+9=0
\end{aligned}
\)
Equation of any plane passing through the intersection of given planes is
\(
\begin{aligned}
& (2 x-y+z-3)+\lambda(4 x-3 y+5 z+9)=0 \\
& \Rightarrow(2+4 \lambda) x+(-1-3 \lambda) y+(1+5 \lambda) z+(-3+9 \lambda)=0 \dots(i)
\end{aligned}
\)
Plane (i) is parallel to given line
\(
\begin{aligned}
& \therefore(-2)(2+4 \lambda)+4(-1-3 \lambda)+5(1+5 \lambda)=0 \\
& \Rightarrow-4-8 \lambda-4-12 \lambda+5+25 \lambda=0 \\
& \Rightarrow 5 \lambda-3=0 \\
& \Rightarrow \lambda=\frac{3}{5}
\end{aligned}
\)
Putting the value of \(\lambda\) in Eq. (i), we get
\(
\begin{aligned}
& (2 x-y+z-3)+\frac{3}{5}(4 x-3 y+5 z+9)=0 \\
& \Rightarrow 11 x-7 y+10 z+6=0 \\
& \therefore a=11, b=-7, c=10 \text { and } d=6 \\
& \therefore a+b+c=11-7+10=14
\end{aligned}
\)

Hint

Equation of \(O P\) is
\(
\begin{aligned}
& \frac{x-0}{3-0}=\frac{y-0}{4-0}=\frac{z-0}{5-0} \\
& \Rightarrow \frac{x}{3}=\frac{y}{4}=\frac{z}{5}
\end{aligned}
\)
Equation of edge parallel to \(Z\)-axis is
\(
\frac{x-3}{0}=\frac{y-0}{0}=\frac{z-5}{1}
\)
\(\therefore\) Shortest distance
\(
=\frac{\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|}{\sqrt{\left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2}}
\)
Here,
\(
\begin{aligned}
& x_1=0, y_1=0, z_1=0 \\
& x_2=3, y_2=0, z_2=5 \\
& a_1=3, b_1=4, c_1=5 \\
& a_2=0, b_2=0, c_2=1 \\
& \therefore\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=\left|\begin{array}{ccc}
3 & 0 & 5 \\
3 & 4 & 5 \\
0 & 0 & 1
\end{array}\right|=4(3)=12 \\
& \text { and } \sqrt{\left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2} \\
& =\sqrt{0+(4-0)^2+(0-3)^2} \\
& =\sqrt{16+9}=5 \\
& \therefore \text { Required shortest distance }=\frac{12}{5} \\
&
\end{aligned}
\)

Hint

Plane \(P\), is passing through intersection of the two planes, so,
\(
\begin{aligned}
& 2 x+3 y-z-2+\lambda(x+2 y+3 z-6)=0 \\
& x(2+\lambda)+y(3+2 \lambda)+z(3 \lambda-1)-2-6 \lambda=0
\end{aligned}
\)
It is perpendicular with plane, \(2 x+y-2+1=0\)
So, \((2+\lambda) 2+(3+2 \lambda) 1+(3 \lambda-1)(-1)=0\)
\(
\lambda=-8
\)
So, plane \(p_1:-6 x-13 y-25 z+46=0\)
Distance of plane \(p\) from the point \((-7,1,1)\)
\(
\begin{aligned}
& d=\frac{|+42-13-25+46|}{\sqrt{36+169+625}}=\frac{50}{\sqrt{830}} \\
& d^2=\frac{2500}{830}=\frac{250}{83}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& L _1: \frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3} \\
& \overrightarrow{a_1}=5 \hat{i}+2 \hat{j}+4 \hat{k} \\
& \overrightarrow{r_1}=\hat{i}+2 \hat{j}-3 \hat{k} \\
& L _2: \frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5} \\
& \overrightarrow{a_2}=-3 \hat{i}-5 \hat{j}+\hat{k} \\
& \overrightarrow{r_2}=\hat{i}+4 \hat{j}-5 \hat{k} \\
& \overrightarrow{r_1} \times \overrightarrow{r_2}=\left|\begin{array}{lll}
1 & \hat{i} & \hat{j} \\
1 & 2 & -3 \\
1 & 4 & -5
\end{array}\right| \\
& =2 \hat{i}+2 \hat{j}+2 \hat{k}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Shortest distance (d) }=\left|\frac{\left(\overrightarrow{r_1} \times \overrightarrow{r_2}\right) \cdot\left(\overrightarrow{a_1}-\overrightarrow{a_2}\right)}{\left|\overrightarrow{r_1} \times \overrightarrow{r_2}\right|}\right| \\
& =\frac{36}{2 \sqrt{3}}=6 \sqrt{3}
\end{aligned}
\)

Hint

Equation of line PM \(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-1}=\lambda\)
Any point on line \(=(\lambda+2,2 \lambda-1,-\lambda+3)\)
For point ‘ \(m\) ‘ \((\lambda+2)+2(2 \lambda-1)-(3-\lambda)=0\)
\(
\begin{aligned}
& \Rightarrow \lambda=\frac{1}{2} \\
& \text { Point } m \left(\frac{1}{2}+2,2 \times \frac{1}{2}-1, \frac{-1}{2}+3\right) \\
& =\left(\frac{5}{2}, 0, \frac{5}{2}\right)
\end{aligned}
\)
For Image \(Q (\alpha, \beta, \gamma)\)
\(
\begin{aligned}
& \frac{\alpha+2}{2}=\frac{5}{2}, \frac{\beta-1}{2}=0 \\
& \frac{\gamma+3}{2}=\frac{5}{2} \\
& Q:(3,1,2) \\
& d=\left|\frac{3(3)+2(1)+2+29}{\sqrt{3^2+2^2+1^2}}\right| \\
& \Rightarrow d=\frac{42}{\sqrt{14}}=3 \sqrt{14}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& P : 8 x+\alpha_1 y +\alpha_2 z +12=0 \\
& L : \frac{ x +2}{2}=\frac{ y -3}{3}=\frac{ z +4}{5}
\end{aligned}
\)
\(\because P\) is parallel to \(L\)
\(
\begin{aligned}
& \Rightarrow 8(2)+\alpha_1(3)+5\left(\alpha_2\right)=0 \\
& \Rightarrow 3 \alpha_1+5\left(\alpha_2\right)=-16
\end{aligned}
\)
Also \(y\)-intercept of plane \(P\) is 1
\(
\Rightarrow \alpha_1=-12
\)
And \(\alpha_2=4\)
\(\Rightarrow\) Equation of plane \(P\) is \(2 x -3 y + z +3=0\)
\(\Rightarrow\) Distance of line \(L\) from Plane \(P\) is
\(
=\left|\frac{0-3(6)+1+3}{\sqrt{4+9+1}}\right|=\sqrt{14}
\)

Hint

\(
\begin{aligned}
& (2 \hat{i}+a \hat{j}+4 \hat{ k }) \cdot[( x -2) \hat{ i }+( y – a ) \hat{ j }+( z -4) \hat{ k }]=0 \\
& \Rightarrow 2 x + ay +4 z =20+ a ^2 \\
& \Rightarrow A \equiv\left(\frac{20+ a ^2}{2}, 0,0\right) \\
& B \equiv\left(0, \frac{20+ a ^2}{ a }, 0\right) \\
& C \equiv\left(0,0, \frac{20+ a ^2}{4}\right) \\
& \Rightarrow \text { Volume of tetrahedron } \\
& =\frac{1}{6}\left[\vec{a} \vec{b} \vec{c}\right]
\end{aligned}
\)
\(
\begin{aligned}
& \left.=\frac{1}{6} \overrightarrow{ a } \cdot \overrightarrow{( b } \times \overrightarrow{ c }\right) \\
& \Rightarrow \frac{1}{6}\left(\frac{20+ a ^2}{2}\right) \cdot\left(\frac{20+ a ^2}{ a }\right) \cdot\left(\frac{20+ a ^2}{4}\right)=144 \\
& \Rightarrow\left(20+ a ^2\right)^3=144 \times 48 \times a \\
& \Rightarrow a =2 \\
& \Rightarrow \text { Equation of plane is } 2 x +2 y +4 z =24 \\
& \text { Or } x + y +2 z =12 \\
& \Rightarrow(3,0,4) \quad \text { Not lies on the } \\
& x + y +2 z =12
\end{aligned}
\)

Hint

Equation of Plane :
\(
2(x-1)-3(y+1)-6(z+5)=0
\)
or \(2 x-3 y-6 z=35\)
\(
\Rightarrow \text { Required distance }=\frac{|2(3)-3(-2)-6(2)-35|}{\sqrt{4+9+36}}=5
\)

Hint

Point \(P (\alpha, \beta, \gamma)\) lies on the plane \(2 x+4 y+3 z=5\),
\(
\therefore 2 \alpha+4 \beta+3 \gamma=5 \dots(1)
\)
Given, \(\left(\begin{array}{lll}\alpha & \beta & \gamma\end{array}\right)\left(\begin{array}{ccc}2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8\end{array}\right)=\left(\begin{array}{lll}0 & 0 & 0\end{array}\right)\)
\(
\therefore 2 \alpha+9 \beta+8 \gamma=0 \dots(2)
\)
and \(10 \alpha+3 \beta+4 \gamma=0 \dots(3)\)
and \(8 \alpha+8 \beta+8 \gamma=0 \dots(4)\)
Subtract (4) from (2)
\(
-6 \alpha+\beta=0 \beta=6 \alpha
\)
From equation (4)
\(
\begin{aligned}
& 8 \alpha+48 \alpha+8 \gamma=0 \\
& \gamma=-7 \alpha
\end{aligned}
\)
From equation (1)
\(
\begin{aligned}
& 2 \alpha+24 \alpha-21 \alpha=5 \\
& 5 \alpha=5 \\
& \alpha=1 \\
& \beta=+6, \quad \gamma=-7 \\
& \therefore 6 \alpha+9 \beta+7 \gamma \\
& =6+54-49 \\
& =11
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0 \\
& \frac{x+1}{1}=\frac{y-1}{1}=\frac{z-4}{-1} \\
& \vec{a}_1=5 \hat{i}+\lambda \hat{j}-\lambda \hat{k}, \vec{a}_2=-\hat{i}+\hat{j}+4 \hat{k} \\
& \vec{a}_1-\vec{a}_2=6 \hat{i}+(\lambda-1) \hat{j}-(\lambda+4) \hat{k} \\
& \vec{b}_1=-2 \hat{i}+\hat{k}, \vec{b}=\hat{i}+\hat{j}-\hat{k} \\
& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 0 & 1 \\
1 & 1 & -1
\end{array}\right| \\
& =-\hat{i}-\hat{j}-2 \hat{k} \\
& \left(\vec{a}_1-\vec{a}_2\right) \cdot \vec{b}_1 \times \vec{b}_2=-6+1-\lambda+2 \lambda+8=\lambda+3 \\
& \text { and }\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{6} \\
& \because \frac{|\lambda+3|}{\sqrt{6}}=2 \sqrt{6} \\
& \therefore \lambda=9, \because \lambda \geq 0 \\
& \therefore \quad L: \frac{x-5}{-2}=\frac{y-9}{0}=\frac{z+9}{1}=k \\
&
\end{aligned}
\)
\(
\therefore \quad \alpha=-2 k+5, \beta=9, \gamma=k-9
\)
Here \(k\) is real then
\(
\alpha+2 \gamma=-13 \neq 24 \text {. }
\)
But all other are in terms of \(k\) hence possible.

Hint

\(
\begin{aligned}
& l=\frac{1}{2}, m=\frac{1}{\sqrt{2}}, n=\cos \theta \\
& l^2+m^2+n^2=1 \\
& \Rightarrow \frac{1}{4}+\frac{1}{2}+n^2=1 \Rightarrow n^2=\frac{1}{4} \Rightarrow n=+\frac{1}{2}
\end{aligned}
\)
\(\theta\) is acute \(\therefore n=\frac{1}{2}\)
\(
\begin{aligned}
& \therefore \vec{v}=k\left(\frac{1}{2} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{2} \widehat{k}\right), k \in R \\
& \vec{v}(\vec{a}-\vec{b})=0 \\
& (\sqrt{2}-a) \frac{1}{2}+(-1-b) \frac{1}{\sqrt{2}}+(1-c) \frac{1}{2}=0 \\
& \Rightarrow \frac{a}{2}+\frac{b}{\sqrt{2}}+\frac{c}{2}=\frac{1}{2} \\
& \Rightarrow a+\sqrt{2} b+c=1
\end{aligned}
\)

Hint

Let \(P \equiv(-1, k, 0), Q \equiv(2, k,-1) \& R(1,1,2)\)
\(
\begin{aligned}
& \overrightarrow{P R}=2 \hat{i}+(1-k) \hat{j}+2 \widehat{k} \\
& \& \vec{Q} R=-\hat{i}+(1-k) \hat{j}+3 \widehat{k}
\end{aligned}
\)
\(\therefore\) Normal to plane will be
\(
\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
2 & (1-k) & 2 \\
-1 & (1-k) & 3
\end{array}\right|=\hat{i}(1-k)-\hat{j}(8)+3 \widehat{k}(1-k)
\)
If line is parallel to this we have
\(
\begin{aligned}
& 1(1-k)+1(-8)+(-3)(1-k)=0 \\
& \Rightarrow 2(1-k)=-8 \\
& \Rightarrow 1-k=-4 \Rightarrow k=5 \\
& \therefore \frac{k^2+1}{(k-1)(k-2)}=\frac{26}{4.3}=\frac{13}{6}
\end{aligned}
\)

Hint

Equation of \(l_1=\frac{x-2}{2}=\frac{y-6}{1}=\frac{z-2}{-2}\)
Shortest distance with \(\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}\) is
\(
\begin{aligned}
& =9 \text { units } \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& C:\left(\frac{a+4}{3}, \frac{-2+2 b}{3}, \frac{4-6}{3}\right) \\
& =\left(\frac{a+4}{3}, \frac{-2+2 b}{3}, \frac{-2}{3}\right)
\end{aligned}
\)
\(
\begin{aligned}
& \text { C lies on plane } 2 x-y+z=4 \\
& 2\left(\frac{a+4}{3}\right)-\left(\frac{2 b-2}{3}\right)-\frac{2}{3}=4 \\
& \Rightarrow 2 a+8-2 b+2-2=12 \\
& \Rightarrow a-b=2 \ldots . \text { (i) }
\end{aligned}
\)
Now, \(O P=\sqrt{5}\)
\(
\begin{aligned}
& \left(\frac{a+4}{3}\right)^2+\left(\frac{2 b-2}{3}\right)^2+\frac{4}{9}=5 \text { and using (i) } \\
& a=\frac{11}{5}, 1 \\
& \Rightarrow b=\frac{1}{5},-1 \\
& \text { as also } \Rightarrow a=1, b=-1 \\
& P(2,-1,-3), c\left(\frac{5}{3}, \frac{-4}{3}, \frac{-2}{3}\right) \\
& C P^2=\frac{1}{9}+\frac{1}{9}+\frac{49}{9}=\frac{17}{3}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}=\lambda \text { (say) } \\
& \& \frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}=\mu(\text { say) } \\
& \therefore \lambda+1=2 \mu+a \ldots \ldots . \text { (i) } \\
& 2 \lambda+2=3 \mu-2 \ldots . . \text { (ii) } \\
& \lambda-3=\mu+3 \ldots . \text { (iii) }
\end{aligned}
\)
By (i) \& (ii)
\(
\begin{aligned}
& \Rightarrow 3 \mu-2=4 \mu+2 a+2 \\
& \mu=-2(1+a) \& \lambda=5-3 a
\end{aligned}
\)
Put \(\lambda \& \mu\) in (iii) we get
\(
\begin{aligned}
& a=-9 \\
& \mu=16 \\
& \lambda=22
\end{aligned}
\)
\(\therefore\) Point of intersection \(\equiv(23,46,19)\)
Distance from \(z=-9\) is 2

Hint

\(
\begin{aligned}
& \overrightarrow{r_1}=\hat{i}-8 \hat{j}+4 \widehat{k} \\
& \overrightarrow{r_2}=\hat{i}+2 \hat{j}+6 \widehat{k} \\
& \vec{a}=2 \hat{i}-7 \hat{j}+5 \widehat{k} \\
& \vec{b}=2 \hat{i}+\hat{j}-3 \widehat{k} \\
& \text { S.D. }=\frac{\left|\overrightarrow{r_1}-\overrightarrow{r_2} \vec{a} \vec{b}\right|}{|\vec{a} \times \vec{b}|}
\end{aligned}
\)
\(
\begin{aligned}
& {\left[\begin{array}{llll}
\overrightarrow{r_1}-\overrightarrow{r_2} & \vec{a} & \vec{b}
\end{array}\right]=\left|\begin{array}{ccc}
0 & -10 & -2 \\
2 & -7 & 5 \\
2 & 1 & -3
\end{array}\right|} \\
& \therefore 10(-16)-2(16)=-192 \\
& \left|\left[\begin{array}{llll}
\overrightarrow{r_1}-\overrightarrow{r_2} & \vec{a} & \vec{b}
\end{array}\right]\right|=192 \\
& \vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
2 & -7 & 5 \\
2 & 1 & -3
\end{array}\right|=16 \hat{i}+16 \hat{j}+16 \widehat{k} \\
& \vec{a} \times \vec{b}=16 \sqrt{3} \\
& \text { S.D. }=\frac{192}{16 \sqrt{3}}=4 \sqrt{3} \\
&
\end{aligned}
\)

Hint

\(
L : \frac{ x +1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}=\lambda \text { (let) }
\)
Let foot of perpendicular is
\(
\begin{aligned}
& P (2 \lambda-1,5 \lambda+1,-\lambda-1) \\
& \overrightarrow{ PA }=(3-2 \lambda) \hat{ i }-(5 \lambda+1) \hat{ j }+(6+\lambda) \hat{ k } \\
& \text { Direction ratio of line } \Rightarrow \overrightarrow{ b }=2 \hat{ i }+5 \hat{ j }-\hat{ k }
\end{aligned}
\)
Now, \(\Rightarrow \overrightarrow{ PA } \cdot \overrightarrow{ b }=0\)
\(
\begin{aligned}
& \Rightarrow 2(3-2 \lambda)-5(5 \lambda+1)-(6+\lambda)=0 \\
& \Rightarrow \lambda=\frac{-1}{6} \\
& P (2 \lambda-1,5 \lambda+1,-\lambda-1) \equiv P (\alpha, \beta, \gamma) \\
& \Rightarrow \alpha=2\left(-\frac{1}{6}\right)-1=-\frac{4}{3} \Rightarrow \alpha=-\frac{4}{3} \\
& \Rightarrow \beta=5\left(-\frac{1}{6}\right)+1=\frac{1}{6} \Rightarrow \beta=\frac{1}{6} \\
& \Rightarrow \gamma=-\lambda-1=\frac{1}{6}-1 \Rightarrow \gamma=-\frac{5}{6}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& L_1: \frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{1}{12}} \\
& L_2: \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}} \\
& \text { S.D }=\left|\frac{(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})}{7}\right| \\
& =\left|\frac{-2-6-6}{7}\right|=2 \text { units }
\end{aligned}
\)

Hint

Equation of line is \(\frac{x+3}{3}\)
\(
\begin{aligned}
= & \frac{y-2}{3}=\frac{z-3}{-1}=\lambda \\
& (3 \lambda-3,3 \lambda+2,3-\lambda)
\end{aligned}
\)
D.R of \(PM (3 \lambda-7,3 \lambda-4,5-\lambda)\)
Since \(P M\) is perpendicular to line
\(
\begin{aligned}
& \Rightarrow 3(3 \lambda-7)+3(3 \lambda-4)-1(5-\lambda)=0 \\
& \Rightarrow \lambda=2 \\
& \Rightarrow M (3,8,1) \Rightarrow PM =\sqrt{14}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Let, } \\
& P \equiv(2 \lambda+1, \lambda+3,2 \lambda+2) \text { and } Q(\mu+2,2 \mu+2,3 \mu+3) \\
& \text { d.r’s of } P Q \equiv<2 \lambda-\mu-1, \lambda-2 \mu+1,2 \lambda-3 \mu-1> \\
& \therefore \quad \frac{2 \lambda-\mu-1}{1}-\frac{\lambda-2 \mu-1}{-1}=\frac{2 \lambda-3 \mu-1}{-2} \\
& \therefore \quad-2 \lambda+\mu+1=\lambda-2 \mu+1 \text { and }-2 \lambda+4 \mu-2= \\
& -2 \lambda+3 \mu+1 \\
& \Rightarrow 3 \lambda-3 \mu=0 \text { and } \mu=3 \\
& \therefore \quad \lambda= \pm 3 \text { and } \mu=3 \\
& \therefore \quad P \equiv(7,6,8) \text { and } Q(5,8,12) \\
& \therefore|P O|=\sqrt{2^2+2^2+4^2}=\sqrt{24}=2 \sqrt{6}
\end{aligned}
\)

Hint

Direction of line
\(
\begin{aligned}
\vec{b} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 1 \\
0 & 3 & -1
\end{array}\right| \\
& =\hat{i}(-5)-\hat{j}(-1)+\hat{k}(3) \\
& =-5 \hat{i}+\hat{j}+3 \hat{k}
\end{aligned}
\)
Equation of line
\(
\frac{x-3}{-5}=\frac{y-2}{1}=\frac{z-1}{3}
\)
Let foot of perpendicular be \(=(-5 k+3, k+2,3 k+1)\)
\(
\Rightarrow(-5 k+2)(-5)+(k-7)(1)+(3 k-6)(3)=0
\)
Or \(25 k-10+k-7+9 k-18=0\)
Or \(k=1\)
\(
\alpha+\beta+\gamma=-k+6=5
\)

Hint

Equation of plane passing through point of intersection of \(P 1\) and \(P 2\)
\(
\begin{aligned}
& P _1+ kP _2=0 \\
& ( x +(\lambda+4) y + z -1)+ k (2 x + y + z -2)=0
\end{aligned}
\)
Passing through \((0,1,0)\) and \((1,0,1)\)
\(
\begin{aligned}
& (\lambda+4-1)+ k (1-2)=0 \\
& (\lambda+3)- k =0 \quad \ldots(1)
\end{aligned}
\)
Also passing \((1,0,1)\)
\(
\begin{aligned}
& (1+1-1)+ k (2+1-2)=0 \\
& 1+ k =0 \\
& k =-1
\end{aligned}
\)
put in (1)
\(
\begin{aligned}
& \lambda+3+1=0 \\
& \lambda=-4
\end{aligned}
\)
Then point \((2 \lambda, \lambda,-\lambda)\)
\(
\begin{aligned}
& (-8,-4,4) \\
& d=\left|\frac{-16-4+4-2}{\sqrt{6}}\right| \\
& d=\frac{18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}=3 \sqrt{6}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& A(2,-3,1), B(-1,1,-2), C(3,-4,2) \\
& \overrightarrow{A B}=-3 \hat{i}+4 \hat{j}-3 \hat{k} \quad \overrightarrow{A C}=\hat{i}-\hat{j}+\hat{k} \\
& \vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-3 & 4 & -3 \\
1 & -1 & 1
\end{array}\right|=\hat{i}-\hat{k}
\end{aligned}
\)
Let equation of plane is \(x-z+\lambda=0\) passes through point
\(
A(2,-3,1) \Rightarrow \lambda=-1
\)
Equation of plane is \(x-z-1=0\)
Distance of point \((7,-3,-4)\) from the plane \(x-z-1=0\) is \(5 \sqrt{2}\)

Hint

Given, \(\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}\)
\(
\Rightarrow \frac{x+4}{3}=\frac{y-2}{-4}=\frac{z-3}{12}
\)

Equation of line passing through the point \(P (-1,9,-16)\) and parallel to line \(\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}\) is \(\frac{x+1}{3}=\frac{y-9}{-4}=\frac{z+16}{12}=\lambda\)
Any point on this line \(( A )=(3 \lambda-1,-4 \lambda+9,12 \lambda-16)\)
Point of intersection line and plane
\(
\begin{aligned}
& 2(3 \lambda-1)+3(-4 \lambda+9)-1(12 \lambda-16)=5 \\
\Rightarrow & 6 \lambda-2-12 \lambda+27-12 \lambda+16=5 \\
\Rightarrow & \lambda=2 \\
\therefore & \text { Point A }=(5,1,8) \\
\therefore & \text { Distance (AP) } \\
= & \sqrt{(5+1)^2+(9-1)^2+(-16-8)^2} \\
= & \sqrt{36+64+576} \\
= & 26 \text { units }
\end{aligned}
\)

Hint

Length of perpendicular
\(
\begin{aligned}
& P Q=\left|\frac{1+4+3-14}{\sqrt{6}}\right|=\sqrt{6} \\
& Q R=(P Q) \cot 60^{\circ}=\sqrt{2}
\end{aligned}
\)
\(\therefore\) Area of \(\triangle PQR =\frac{1}{2}( PQ )( QR )=\sqrt{3}\)

Hint

\(\because A(2,3,9), B(5,2,1), C(1, \lambda, 8)\) and \(D (\lambda, 2,3)\) are coplanar.
\(
\therefore\left[\begin{array}{lll}
\overrightarrow{ AB } & \overrightarrow{ AC } & \overrightarrow{ AD }
\end{array}\right]=0
\)
\(
\begin{aligned}
& \left|\begin{array}{ccc}
3 & -1 & -8 \\
-1 & \lambda-3 & -1 \\
\lambda-2 & -1 & -6
\end{array}\right|=0 \\
& \Rightarrow[-6(\lambda-3)-1]-8(1-(\lambda-3)(\lambda-2))+(6+(\lambda-2)=0 \\
& \Rightarrow 3(-6 \lambda+17)-8\left(-\lambda^2+5 \lambda-5\right)+(\lambda+4)=8 \\
& \Rightarrow 8 \lambda^2-57 \lambda+95=0 \\
& \therefore \lambda_1 \lambda_2=\frac{95}{8}
\end{aligned}
\)

Hint

\(\left(-2, \frac{7}{2}, \frac{3}{2}\right)\) satisfies the plane \(P: 2 x+m y+n z=4\)
\(
-4+\frac{7 m}{2}+\frac{3 n}{2}=4 \Rightarrow 7 m+3 n=16 \dots(i)
\)
Line joining \(A(-1,4,3)\) and \(\left(-2, \frac{7}{2}, \frac{3}{2}\right)\) is perpendicular to
\(
\begin{aligned}
& P: 2 x+m y+n z=4 \\
& \frac{1}{2}=\frac{\frac{1}{2}}{m}=\frac{\frac{3}{2}}{n} \Rightarrow m=1 \& n=3
\end{aligned}
\)
Plane \(P: 2 x+y+3 z=4\)
Distance of \(P\) from \(A(-1,4,3)\) parallel to the line
\(
\frac{x+1}{3}=\frac{y-4}{-1}=\frac{z-3}{-4}: L
\)
for point of intersection of \(P \& L\)
\(
2(3 r-1)+(-r+4)+3(-4 r+3)=4 \Rightarrow r=1
\)
Point of intersection : \((2,3,-1)\)
Required distance \(=\sqrt{3^2+1^2+4^2}=\sqrt{26}\)

Hint

\(
L_1: \frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}
\)
through a point \(\vec{a}_1 \equiv(1,2,3)\)
parallel to \(\overrightarrow{b_1} \equiv(\lambda, 1,2)\)
\(
L_2: \frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}
\)
through a point \(\vec{a}_2=(-26,-18,-28)\)
parallel to \(\overrightarrow{b_2}=(-2,3,1)\)
If lines are coplanar then, \(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot \overrightarrow{b_1} \times \overrightarrow{b_2}=0\)
\(
\Rightarrow\left|\begin{array}{ccc}
27 & 20 & 31 \\
\lambda & 1 & 2 \\
-2 & 3 & \lambda
\end{array}\right|=0 \Rightarrow \lambda=3
\)
Vector normal to the required plane \(\vec{n}=\overrightarrow{b_1} \times \overrightarrow{b_2}\)
\(
\Rightarrow \vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
3 & 1 & 2 \\
-2 & 3 & 3
\end{array}\right|=-3 \hat{i}-13 \hat{j}+11 \widehat{k}
\)
Equation of plane
\(
\begin{aligned}
& \equiv((x-1),(y-2),(z-3)) \cdot(-3,-13,11)=0 \\
& \Rightarrow 3 x+13 y-11 z+4=0
\end{aligned}
\)
Checking the option gives \((0,4,5)\) does not lie on the plane.

Hint

Normal of plane P:
\(
\begin{aligned}
\vec{n} & =\left|\begin{array}{ccc}
i & j & k \\
-2 & 1 & -3 \\
-1 & 2 & -2
\end{array}\right| \\
& =4 i – j -3 k
\end{aligned}
\)
Equation of plane P which passes through (2,2,-2) is
\(
\begin{gathered}
4( x -2)-1( y -2)-3( z +2)=0 \\
4 x -8- y +2-3 z -6=0 \\
4 x – y -3 z =12 \\
\frac{ x }{3}+\frac{ y }{-12}+\frac{ z }{(-4)}=1 \\
\alpha=3, \beta=-12,\gamma=-4 \\
\alpha+\beta+\gamma=-13 \\
\text { Volume }=\frac{1}{6}[\alpha \times \beta] \times \gamma\\
=\frac{1}{6} \times 3 \times 12 \times-4 \\
=24
\end{gathered}
\)

Hint

Any point on \(x^2+y^2=1, z=0\) is \(p(\cos \theta, \sin \theta, 0)\)
If foot of perpendicular of \(p\) on the plane \(2 x+3 y+z=6\) is \((h, k, l)\) then
\(
\begin{aligned}
& \frac{h-\cos \theta}{2}=\frac{k-\sin \theta}{3}=\frac{l-0}{1} \\
& =-\left(\frac{2 \cos \theta+3 \sin \theta+0-6}{2^2+3^2+1^2}\right)=r \text { (let) } \\
& h=2 r+\cos \theta, k=3 r+\sin \theta, l=r
\end{aligned}
\)
Hence, \(h-2 l=\cos \theta\) and \(k-3 l=\sin \theta\)
Hence, \((h-2 l)^2+(k-3 l)^2=1\)
When \(l=6-2 h-3 k\)
Hence required locus is
\(
\begin{aligned}
& (x-2(6-2 x-3 y))^2+(y-3(6-2 x-3 y))^2=1 \\
& \Rightarrow(5 x+6 y-12)^2+4(3 x+5 y-9)^2=1, z=6-2 x-3 y
\end{aligned}
\)

Hint

\(PR\) is perpendicular to given line, so
\(
\begin{aligned}
& 2(2 \lambda-1-a)+3(3 \lambda-1)-1(-\lambda-1)=0 \\
& \Rightarrow a=7 \lambda-2
\end{aligned}
\)
Now,
\(
\begin{aligned}
& \because P R=2 \sqrt{6} \\
& \Rightarrow(-5 \lambda+1)^2+(3 \lambda-1)^2+(\lambda+1)^2=24 \\
& \Rightarrow 5 \lambda^2-2 \lambda-3=0 \Rightarrow \lambda=1 \text { or }-\frac{3}{5} \\
& \because a>0 \text { so } \lambda=1 \text { and } a=5 \\
& \text { Now } \left.\sum_{i=1}^3 \alpha_i=2 \text { (Sum of co-ordinate of } R \right)-(\text { Sum of coordinates of } P ) \\
& =2(7)-11=3 \\
& a+\sum_{i=1}^3 \alpha_i=5+3=8
\end{aligned}
\)

Hint

\(
\begin{aligned}
& P_1: a x+b y+0 z=3, \text { normal vector }: \vec{n}_1=(a, b, 0) \\
& P_2: a x+b y+c z=0, \text { normal vector }: \vec{n}_2=(a, b, c) \\
& \text { Vector parallel to the line of intersection }=\vec{n}_1 \times \vec{n}_2 \\
& \vec{n}_1 \times \vec{n}_2=(b c,-a c, 0)
\end{aligned}
\)
Vector normal to \(0 . x+y-z+2=0\) is \(\vec{n}_3=(0,1,-1)\)
Angle between line and plane is \(30^{\circ}\)
\(
\begin{aligned}
& \Rightarrow\left|\frac{0-a c+0}{\sqrt{b^2 c^2+c^2 a^2} \sqrt{2}}\right|=\frac{1}{2} \\
& \Rightarrow a^2=b^2
\end{aligned}
\)
Hence, \(\vec{n}_1 \times \vec{n}_2=(a c,-a c, 0)\)
Direction ratios \(=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, 0\right)\)

Hint

\(
\begin{aligned}
& P_1: 2 x+k y-5 z=1 \\
& P_2: 3 k x-k y+z=5 \\
& \because P_1 \perp P_2 \Rightarrow 6 k-k^2+5=0 \\
& \Rightarrow k=1,5 \\
& \because k<3 \\
& \therefore k=1 \\
& P_1: 2 x+y-5 z=1 \\
& P_2: 3 x-y+z=5 \\
& P:(2 x+y-5 z-1)+\lambda(3 x-y+z-5)=0 \\
& \text { Positive x-axis intercept }=1 \\
& \Rightarrow \frac{1+5 \lambda}{2+3 \lambda}=1 \\
& \Rightarrow \lambda=\frac{1}{2} \\
& \therefore P: 7 x+y-4 z=7 \\
& \text { y intercept }=7 .
\end{aligned}
\)

Hint

If \(\vec{n}_1\) is a vector normal to the plane determined by \(\hat{i}\) and \(\hat{i}+\hat{j}\) then
\(
\vec{n}_1=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
1 & 0 & 0 \\
1 & 1 & 0
\end{array}\right|=\widehat{k}
\)
If \(\overrightarrow{n_2}\) is a vector normal to the plane determined by \(\hat{i}-\hat{j}\) and \(\hat{i}+\hat{k}\) then
\(
\bar{n}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
1 & -1 & 0 \\
1 & 0 & 1
\end{array}\right|=-\hat{i}-\hat{j}+\widehat{k}
\)
Vector \(\vec{a}\) is parallel to \(\vec{n}_1 \times \vec{n}_2\)
i.e. \(\vec{a}\) is parallel to \(\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 0 & 0 & 1 \\ -1 & -1 & 1\end{array}\right|=\hat{i}-\hat{j}\)
Given \(\vec{b}=\hat{i}-2 \hat{j}+2 \widehat{k}\)
Cosine of acute angle between
\(
\vec{a} \text { and } \vec{b}=\left|\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}\right|=\frac{1}{\sqrt{2}}
\)
Obtuse angle between \(\vec{a}\) and \(\vec{b}=\frac{3 \pi}{4}\)

Hint

The line \(x+y-z=0=x-2 y+3 z-5\) is parallel to the vector
\(
\vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
1 & 1 & -1 \\
1 & -2 & 3
\end{array}\right|=(1,4,-3)
\)
Equation of the line through \(P(1,2,4)\) and parallel to \(\vec{b}\)
\(
\frac{x-1}{1}=\frac{y-2}{-4}=\frac{z-4}{-3}
\)
Let \(N \equiv(\lambda+1,-4 \lambda+2,-3 \lambda+4)\)
\(
\overrightarrow{Q N}=(\lambda,-4 \lambda+4,-3 \lambda-1)
\)
\(\overrightarrow{Q N}\) is perpendicular to \(\vec{b}\)
\(
\begin{aligned}
& \Rightarrow(\lambda,-4 \lambda+4,-3 \lambda-1) \cdot(1,4,-3)=0 \\
& \Rightarrow \lambda=\frac{1}{2}
\end{aligned}
\)
Hence \(\overrightarrow{Q N}=\left(\frac{1}{2}, 2, \frac{-5}{2}\right)\) and \(|\overrightarrow{Q N}|=\sqrt{\frac{21}{2}}\)

Hint

Let equation of plane be \(a(x-1)+b(y+1)+c(z-1)=0\)…..(1)
It is perpendicular to the given two planes
\(
\begin{aligned}
& 2 a-2 b+c=0 \\
& a-b+2 c=0 \\
& \Rightarrow \frac{a}{3}=\frac{b}{3}=\frac{c}{0}
\end{aligned}
\)
Equation of plane be \(x+y=0\)
Now \(\frac{|a+a|}{\sqrt{2}}=3 \sqrt{2} \Rightarrow|2 a|=6 \Rightarrow a= \pm 3\)
\(
\begin{aligned}
& P(3,3,2) \text { or } P(-3,-3,2), Q(1,-1,1) \\
& P Q^2=(3-1)^2+(3+1)^2+(2-1)^2=21
\end{aligned}
\)

Hint

\(
L_1: \frac{x+7}{6}=\frac{y-6}{7}=\frac{z-0}{1}
\)
Any point on it \(\vec{a}_1(-7,6,0)\) and \(L _1\) is parallel to \(\vec{b}_1(-6,7,1)\) \(L_2: \frac{x-7}{-2}=\frac{y-2}{1}=\frac{z-6}{1}\)
Any point on it \(\vec{a}_2(7,2,6)\) and \(L _2\) is parallel to \(\vec{b}_2(-2,1,1)\)
Shortest distance between \(L _1\) and \(L _2\)
\(
=\left|\frac{\left.\left.\overrightarrow{\left(a_2\right.}-\overrightarrow{a_1}\right) \cdot \overrightarrow{\left(b_1\right.} \times \overrightarrow{b_2}\right)}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\right|
\)
\(
=\left|\frac{(-14,4,-6) \cdot(3,2,4)}{\sqrt{9+4+16}}\right|
\)
\(
=2 \sqrt{29} .
\)

Hint

Let \(\langle a, b, c\rangle\) be direction ratios of plane containing lines \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\) and \(\frac{x}{3}=\frac{y}{7}=\frac{z}{8}\).
\(
\therefore \quad 2 a+3 b+5 c=0 \quad \ldots \text { (i) }
\)
and \(3 a+7 b+8 c=0 \quad \ldots(ii)\)
from eq. (i) and (ii) : \(\frac{a}{24-35}=\frac{b}{15-16}=\frac{c}{14-9}\)
\(\therefore\) D. R’s. of plane are \(<11,1,-5>\)
Let \(D\). R’s of plane \(P\) be \(<a_1, b_1, c_1>\) then.
\(
11 a_1+b_1-5 c_1=0 \dots(iii)
\)
and \(9 a_1-b_1-5 c_1=0 \dots(iv)\)
From eq. (iii) and (iv) :
\(
\frac{a_1}{-5-5}=\frac{b_1}{-45+55}=\frac{c_1}{-11-9}
\)
\(\therefore\) D.R’s of plane \(P\) are \(\langle 1,-1,2\rangle\)
Equation plane \(P\) is : \(1(x-3)-1(y+4)+2(z-7)=0\)
\(
\Rightarrow x-y+2 z-21=0
\)
Distance from point \((2,-5,11)\) is \(d=\frac{|2+5+22-2|}{\sqrt{6}}\)
\(
\therefore d^2=\frac{32}{3}
\)

Hint

Line \(PQ\) is parallel to line \(\frac{2-x}{2}=\frac{y-3}{2}=\frac{z+1}{1}\)
\(\therefore D R\) of \(P Q=D R\) of line \(=<-2,2,1>\)
\(\therefore\) Equation of line \(P Q\) passing through \(P(3,2,-1)\) and \(D R=<-2,2,1\rangle\) is \(\frac{x-3}{-2}=\frac{y-2}{2}=\frac{z+1}{1}\)
Any General point on line \(P Q=\left(x_1, y_1, z_1\right)\)
\(
\begin{aligned}
& \therefore \frac{x_1-3}{-2}=\frac{y_1-2}{2}=\frac{z_1+1}{1}=\lambda \\
& \Rightarrow x_1=-2 \lambda+3 \\
& y_1=2 \lambda+2 \\
& z_1=\lambda-1
\end{aligned}
\)
\(
\therefore \text { Point } Q =(-2 \lambda+3,2 \lambda+2, \lambda-1)
\)
Point \(Q\) lies on the plane \(3 x-y+4 z+1=0\). So point \(Q\) satisfy the equation.
\(
\begin{aligned}
& 3(-2 \lambda+3)-(2 \lambda+2)+4(\lambda-1)+1=0 \\
& \Rightarrow-6 \lambda+9-2 \lambda-2+4 \lambda-4+1=0 \\
& \Rightarrow-4 \lambda+4=0 \\
& \Rightarrow \lambda=1 \\
& \therefore \text { Point } Q(-2 \times 1+3,2 \times 1+2,1-1) \\
& =(1,4,0)
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \text { Distance of the point } P (3,2,-1) \text { from the plane }=\text { Length of } PQ \\
& =\sqrt{(3-1)^2+(2-4)^2+(-1-0)^2} \\
& =\sqrt{2^2+(-2)^2+(-1)^2} \\
& =\sqrt{4+4+1} \\
& =\sqrt{9} \\
& =3
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}=\lambda \\
& (3 \lambda+2,-2 \lambda-1,-\lambda-3) \text { lies on plane } p x-q y+z=5 \\
& p(3 \lambda+2)-q(-2 \lambda-1)+(-\lambda-3)=5 \\
& \lambda(3 p+2 q-1)+(2 p+q-8)=0 \\
& 3 p+2 q-1=0\} p=15 \\
& 2 p+q-8=0\} q=-22
\end{aligned}
\)
Equation of plane \(15 x+22 y+z-5=0\)
Shortest distance from origin \(=\frac{|0+0+0-5|}{\sqrt{15^2+22^2+1}}\)
\(
\begin{aligned}
& =\frac{5}{\sqrt{710}} \\
& =\sqrt{\frac{5}{142}}
\end{aligned}
\)

Hint

Given: Plane: \(x+2 y+2 z-16=0\)
And line: \(\vec{r}=-\widehat{k}+\lambda(\hat{i}+\hat{j}+2 \widehat{k})\)
So, equation of line in symmetric form is \(\frac{x-0}{1}=\frac{y-0}{1}=\frac{z+1}{2}\)
Now, mirror image of \(p(1,2,1)\) in plane \(x+2 y+2 z-16=0\) is
\(
\begin{aligned}
& \Rightarrow \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=-2\left(\frac{1+2(2)+2(1)-16}{1^2+2^2+2^2}\right) \\
& \Rightarrow \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=2 \\
& \Rightarrow x =3, y =6, z =5
\end{aligned}
\)
\(\therefore\) Coordinates of point \(Q\) are \((3,6,5)\)
Now, equation of plane \(T\) is \(\left|\begin{array}{ccc}x & y & z+1 \\ 1 & 1 & 2 \\ 3 & 6 & 6\end{array}\right|=0\)
\(
\begin{aligned}
& \Rightarrow x(6-12)-y(6-6)+(z+1)(6-3)=0 \\
& \Rightarrow-6 x+3 z+3=0 \\
& \Rightarrow 2 x-z-1=0
\end{aligned}
\)
So, \((1,2,1)\) lies on plane \(T\).

Hint

We know mirror image of point \(\left(x_1, y_1, z_1\right)\) in the plane \(a x+b y+c z=d\)
\(
\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-2\left(a x_1+b y_1+c z_1-d\right)}{a^2+b^2+c^2}
\)
Here given point \((2,4,7)\) and plane \(3 x-y+4 z=2\) then mirror image is
\(
\begin{aligned}
& \frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-2(6-4+28-2)}{9+1+16} \\
& \Rightarrow \frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=-\frac{28}{13} \\
& \therefore x=-\frac{58}{13}=a \\
& y=\frac{80}{13}=b \\
& z=-\frac{21}{13}=c \\
& \therefore 2 a+b+2 c \\
& =2\left(-\frac{58}{13}\right)+\frac{80}{13}+2\left(-\frac{21}{13}\right) \\
& =\frac{-116+80-42}{13}=\frac{-78}{13}=-6
\end{aligned}
\)

Hint

Equation of pane through point \((2,3,-5)\) and perpendicular to planes \(2 x+y-5 z=10\) and \(3 x+5 y-7 z=12\) is
\(
\left|\begin{array}{ccc}
x-2 & y-3 & z+5 \\
2 & 1 & -5 \\
3 & 5 & -7
\end{array}\right|=0
\)
\(\therefore\) Equation of plane is \((x-2)(-7+25)-(y-3)\)
\(
\begin{aligned}
& (-14+15)+(z+5) \cdot 7=0 \\
& \therefore 18 x-y+7 z+2=0 \\
& \Rightarrow 18 x-y+7 z=-2 \\
& \therefore-18 x+y-7 z=2
\end{aligned}
\)
On comparing with \(a x+b y+c z=d\) where \(d >0\) is \(a =-18, b =1, c =-7, d =2\)
\(
\therefore a+7 b+c+20 d=22
\)

Hint

\(
\begin{aligned}
& -x+2 y-z=0 \\
& 3 x-5 y+2 z=0 \\
& \vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 2 & -1 \\
3 & -5 & 2
\end{array}\right| \\
& =\hat{i}(-1)-\hat{j}(1)+\hat{k}(-1) \\
& \vec{n}=-\hat{i}-\hat{j}-\hat{k}
\end{aligned}
\)
Equation of LOI is \(\frac{x}{1}=\frac{y}{1}=\frac{z}{1}\)
DR: of \(PT \rightarrow \alpha-1, \alpha+2, \alpha-3\)
\(DR\) : of \(QR \rightarrow \quad 1, \quad 1,1\)
\(
\begin{aligned}
& \Rightarrow(\alpha-1) \times 1+(\alpha+2) \times 1+(\alpha-3) \times 1=0 \\
& 3 \alpha=2 \\
& \alpha=\frac{2}{3}
\end{aligned}
\)
\(
\begin{aligned}
& PT ^2=\frac{1}{9}+\frac{64}{9}+\frac{49}{9} \\
& PT ^2=\frac{114}{9} \\
& PT =\frac{\sqrt{114}}{3}
\end{aligned}
\)
\(
\cos \theta=\frac{\sqrt{114}}{3} \times \frac{1}{3 \sqrt{2}}=\frac{\sqrt{57}}{9}=\frac{\sqrt{19 \times 3}}{3 \times 3}=\frac{\sqrt{19}}{3 \sqrt{3}}
\)
\(
\begin{aligned}
& \cos 2 \theta=\frac{2 \times 19}{27}-1=\frac{11}{27} \\
& \sin 2 \theta=\sqrt{1-\left(\frac{11}{27}\right)^2}=\frac{\sqrt{38} \sqrt{16}}{27}=\frac{4}{27} \sqrt{38}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Area }=\frac{1}{2} \times \sqrt{18} \sqrt{18} \times \frac{4}{27} \sqrt{38} \\
& =\frac{18}{2} \times \frac{4}{27} \sqrt{38}=\frac{36}{27} \sqrt{38}=\frac{4}{3} \sqrt{38}
\end{aligned}
\)

Hint

Family of Plane’s equation can be given by
\(
(5+8 \lambda) x+(8-7 \lambda) y+(13+\lambda) z-(29+20 \lambda)=0
\)
\(P_1\) passes through \((2,1,3)\)
\(
\begin{aligned}
& \Rightarrow(10+16 \lambda)+(8-7 \lambda)+(39+3 \lambda)-(29+20 \lambda)=0 \\
& \Rightarrow-8 \lambda+28=0 \Rightarrow \lambda=\frac{7}{2}
\end{aligned}
\)
d.r, s of normal to \(P _1\)
\(
\left\langle 33, \frac{-33}{2}, \frac{33}{2}\right\rangle \text { or }\left\langle 1,-\frac{1}{2}, \frac{1}{2}\right\rangle
\)
\(P_2\) passes through \((0,1,2)\)
\(
\begin{aligned}
& \Rightarrow 8-7 \lambda+26+2 \lambda-(29+20 \lambda)=0 \\
& \Rightarrow 5-25 \lambda=0 \\
& \Rightarrow \lambda=\frac{1}{5}
\end{aligned}
\)
d.r, \(s\) of normal to \(P_2\)
\(
\left\langle\frac{33}{5}, \frac{33}{5}, \frac{66}{5}\right\rangle \text { or }\langle 1,1,2\rangle
\)
Angle between normals
\(
\begin{aligned}
& =\frac{\left(\hat{i}-\frac{1}{2} \hat{j}+\frac{1}{2} \hat{k}\right) \cdot(\hat{i}+\hat{j}+2 \hat{k})}{\frac{\sqrt{3}}{2}} \\
& \cos \theta=\frac{1-\frac{1}{2}+1}{3}=\frac{1}{2} \\
& \theta=\frac{\pi}{3} \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& P_1: x+3 y-z=6 \\
& P_2:-6 x+5 y-z=7
\end{aligned}
\)
Family of planes passing through line of intersection of \(P_1\) and \(P_2\) is given by \(x(1-6 \lambda)+y(3+5 \lambda)+z(-1-\lambda)-(6+7 \lambda)=0\)
It passes through \(\left(2,3, \frac{1}{2}\right)\)
\(
\begin{aligned}
& \text { So, } 2(1-6 \lambda)+3(3+5 \lambda)+\frac{1}{2}(-1-\lambda)-(6+7 \lambda)=0 \\
& \Rightarrow 2-12 \lambda+9+15 \lambda-\frac{1}{2}-\frac{\lambda}{2}-6-7 \lambda=0 \\
& \Rightarrow \frac{9}{2}-\frac{9 \lambda}{2}=0 \Rightarrow \lambda=1
\end{aligned}
\)
Required plane is
\(
\begin{aligned}
& -5 x+8 y-2 z-13=0 \\
& \text { Or } \vec{r} \cdot(-5 \hat{i}+8 \hat{j}-2 \widehat{k})=13 \\
& \frac{|13 \vec{a}|^2}{|d|^2}=\frac{13^2}{(13)^2} \cdot|\vec{a}|^2=93
\end{aligned}
\)

Hint

\(
L: \frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}=t
\)
Let \(P=(4 t-2,2 t+1,3 t-1)\)
\(\because P\) is the foot of perpendicular of \((1,2,4)\)
\(
\begin{aligned}
& \therefore 4(4 t-3)+2(2 t-1)+3(3 t-5)=0 \\
& \Rightarrow 29 t=29 \Rightarrow t=1 \\
& \therefore P =(2,3,2)
\end{aligned}
\)
Now, distance of \(P\) from the plane \(3 x+4 y+12 z+23=0\), is
\(
\left|\frac{6+12+24+23}{\sqrt{9+16+144}}\right|=\frac{65}{13}=5
\)

Hint

\(
\begin{aligned}
& L_1: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1} \\
& L_2: \frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3} \\
& \text { Now, } \vec{p} \times \vec{q}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & -1 \\
2 & 1 & 3
\end{array}\right|=10 \hat{i}-8 \hat{j}-4 \widehat{k} \\
& \text { and } \overrightarrow{a_2}-\overrightarrow{a_1}=6 \hat{i}-4 \hat{j}-4 \widehat{k} \\
& \therefore S . D .=\left|\frac{60+32+16}{\sqrt{100+64+16}}\right|=\frac{108}{\sqrt{180}}=\frac{18}{\sqrt{5}}
\end{aligned}
\)

Hint

Given that the direction cosines satisfy \(l+m-n=0\), we find that \(n=l+m\).
The other equation is \(3 l^2+m^2+c n l=0\), and substituting \(n=l+m\) gives \(3 l^2+m^2+c l(l+m)=0\).
This simplifies to \((3+c) l^2+c l m+m^2=0\).
As the lines are parallel, they share the same direction ratios, so we can express \(l\) in terms of \(m\), say \(l=k m\). Substituting this into our equation gives
\(
(3+c)(k m)^2+c k m^2+m^2=0 \text {. }
\)
This simplifies to \(m^2\left[k^2(3+c)+k c+1\right]=0\).
Since \(m \neq 0\), we must have \(k^2(3+c)+k c+1=0\). Here, we consider the ratio \(k=\frac{l}{m}\) to be constant, since the lines are parallel. The equation then becomes a quadratic equation in \(k\).
As the lines are parallel, the discriminant of the quadratic equation must be equal to zero for the equation to have equal roots. Hence, the discriminant
\(
D=\left(c^2-4(3+c)\right)=0 \text {. }
\)
Solving this quadratic equation gives \(c^2-4 c-12=0\).
Factoring this equation gives \((c-6)(c+2)=0\).
Solving for \(c\) gives \(c=6,-2\).
However, we are looking for the positive value of \(c\), so \(c=6\).
Therefore, the correct answer is 6.

Hint

\(
\begin{aligned}
& P_1: 2 x+y-52=0, P_2: 3 x-y+4 z-7=0 \\
& \text { Family of planes } P _1 \text { and } P _2 \\
& P: P_1+\lambda P_2 \\
& \therefore P:(2+3 \lambda) x+(1-\lambda) y+(-5+4 \lambda) z-7 \lambda=0 \\
& \because P \perp P_1 \\
& \therefore 4+6 \lambda+1-\lambda+25-20 \lambda=0 \\
& \lambda=2 \\
& \therefore P: 8 x-y+32-14=0
\end{aligned}
\)
It passes through the point \((1,0,2)\)