Past JEE Main Entrance Papers Set-II

Hint

Line is \(\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}=\lambda\) : Let point \(R\) is
\(
(2 \lambda-1,-2 \lambda+3,-\lambda)
\)
Direction ratio of \(P Q \equiv(2 \lambda-2,-2 \lambda+1,3-\lambda)\)
\(P O\) is \(\perp^r\) to line
\(
\begin{aligned}
& \Rightarrow 2(2 \lambda-2)-2(-2 \lambda+1)-1(3-\lambda)=0 \\
& 4 \lambda-4+4 \lambda-2-3+\lambda=0 \\
& 9 \lambda=9 \Rightarrow \lambda=1 \\
& \Rightarrow \text { Point } R \text { is }(1,1,-1) \\
& \begin{array}{c|c|c}
\frac{a+1}{2}=1 & \frac{b+2}{2}=1 & \frac{c-3}{2}=-1 \\
a=1 & b=0 & c=1
\end{array} \\
& \Rightarrow a+b+c=2 \\
&
\end{aligned}
\)

Hint

Equation of line parallel to \(\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}\) passes through \((1,-2,3)\) is
\(
\begin{aligned}
& \frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r \\
& x=2 r+1 \\
& y=3 r-2 \\
& z=-6 r+3
\end{aligned}
\)
A point on whole line \(=(2 r+1,3 r-2,-6 r+3)\).
This point lies on plane \(x-y+2=5\)
\(
\begin{aligned}
& \text { so, } 2 r+1-3 r+2-6 r+3=5 \\
& \Rightarrow r=\frac{1}{7} \\
& \therefore x=\frac{9}{7}, y=\frac{-11}{7}, z=\frac{15}{7}
\end{aligned}
\)
Distance is \(=\sqrt{\left(\frac{9}{7}-1\right)^2+\left(2-\frac{11}{7}\right)^2+\left(3-\frac{15}{7}\right)^2}\)
\(
\begin{aligned}
& =\sqrt{\left(\frac{2}{7}\right)^2+\left(\frac{3}{7}\right)^2+\left(\frac{6}{7}\right)^2} \\
& =\frac{1}{7} \sqrt{4+9+36}=1
\end{aligned}
\)

Hint

Direction ratios of normal to plane are <2,-6,4>
Also plane passes through \((3,1,1)\)
\(\therefore\) Equation of plane
\(
\begin{aligned}
& 2( x -3)-6( y -1)+4( z -1)=0 \\
& \Rightarrow x -3 y +2 z =2
\end{aligned}
\)
By checking all options we can see this equation passes through \((4,0,-1)\)

Hint

Equation of \(A B\),
\(
\begin{aligned}
& \frac{x-1}{0}=\frac{y+2}{3}=\frac{z-3}{-3}=\lambda \\
& \therefore \text { Coordinates of any point on the line }( M )=(-3,3 \lambda-2,-3 \lambda) \\
& \overrightarrow{P M}=-3 \hat{i}+(3 \lambda-4) \hat{j}-3 \lambda \widehat{k} \\
& \overrightarrow{A B}=3 \hat{j}-3 \widehat{k}
\end{aligned}
\)
As \(\overrightarrow{P M} \perp \overrightarrow{A B}\)
\(
\begin{aligned}
& \therefore \overrightarrow{P M} \cdot \overrightarrow{A B}=0 \\
& \Rightarrow(-3) \cdot 0+(3 \lambda-4)(3)+(-3 \lambda)(-3)=0 \\
& \Rightarrow \lambda=\frac{2}{3} \\
& \therefore M=(1,0,1)
\end{aligned}
\)
By checking each options we can see \(M\) lies on \(2 x+y-z=1\).

Hint

As normal is perpendicular to both the lines so normal vector to the plane is
\(
\begin{aligned}
& \vec{n}=(\hat{i}-2 \hat{j}+2 \widehat{k}) \times(2 \hat{i}+3 \hat{j}-\widehat{k}) \\
& \vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
1 & -2 & 2 \\
2 & 3 & -1
\end{array}\right| \\
& \vec{n}=(2-6) \hat{i}-(-1-4) \hat{j}+(3+4) \widehat{k} \\
& \vec{n}=-4 \hat{i}+5 \hat{j}+7 \widehat{k}
\end{aligned}
\)
Now equation of plane passing through \((3,1,1)\) is
\(
\begin{aligned}
& \Rightarrow-4(x-3)+5(y-1)+7(z-1)=0 \\
& \Rightarrow-4 x+12+5 y-5+7 z-7=0 \\
& \Rightarrow-4 x+5 y+7 z=0 \dots(1)
\end{aligned}
\)
Plane is also passing through \((\alpha,-3,5)\) so this point satisfies the equation of plane so put in equation (1)
\(
\begin{aligned}
& -4 \alpha+5 \times(-3)+7 \times(5)=0 \\
& \Rightarrow-4 \alpha-15+35=0 \\
& \Rightarrow \alpha=5
\end{aligned}
\)

Hint

Equation of plane passing through \((2,1,2)\)
\(
a(x-2)+b(y-1)+c(z-2)=0
\)
As point \((1,2,1)\) also passes through the plane, so it satisfy the equation,
\(
\begin{aligned}
& a(1-2)+b(2-1)+c(1-2)=0 \\
& \Rightarrow-a+b-c=0 \ldots .(2)
\end{aligned}
\)
Given line \(2 x=3 y\) and \(z=1\),
So, symmetric form of the line
\(
\frac{x}{3}=\frac{y}{2}=\frac{z-1}{0}
\)
\(\therefore\) Direction ratio of this line is \((3,2,0)\) and Direction ration of plane \(=(a, b, c)\)
As plane is parallel to the line so the normal of the plane is perpendicular to the line.
\(\therefore\) Dot product of direction ratio \(=0\)
\(
3 a+2 b+0(c)=0
\)
Equation of plane,
\(
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-1 & z-2 \\
-1 & 1 & -1 \\
3 & 2 & 0
\end{array}\right|=0 \\
& \Rightarrow 3(1-y+2-z)-2(-x+2+z-2)=0 \\
& \Rightarrow 9-3 y-3 z+2 x-2 z=0 \\
& \Rightarrow 2 x-3 y-5 z+9=0
\end{aligned}
\)
By checking all options you can see \((-2,0,1)\) satisfy the equation.

Hint

Let \(A (1,2,3), B \left(-\frac{7}{3},-\frac{4}{3},-\frac{1}{3}\right)\)
\(\therefore\) Midpoint of \(A B=M=\left(\frac{\frac{-7}{3}+1}{2}, \frac{\frac{4}{3}+2}{2}, \frac{\frac{-1}{3}+3}{2}\right)\)
\(
=\left(\frac{-2}{3}, \frac{1}{3}, \frac{4}{3}\right)
\)
DR of AM \(=\left(1+\frac{2}{3}, 2-\frac{1}{3}, 3-\frac{4}{3}\right)\)
\(
\begin{aligned}
& =\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right) \\
& =(1,1,1)
\end{aligned}
\)
Equation of plane
\(
\begin{aligned}
& a\left(x+\frac{2}{3}\right)+b\left(y-\frac{1}{3}\right)+c\left(z-\frac{4}{3}\right)=0 \\
& \Rightarrow 1\left(x+\frac{2}{3}\right)+1\left(y-\frac{1}{3}\right)+1\left(z-\frac{4}{3}\right)=0 \\
& \Rightarrow x+y+z=1
\end{aligned}
\)
\(\therefore(1,-1,1)\) lies on the plane.

Hint

\(
\begin{aligned}
& \vec{a}=\langle 3,8,3\rangle \\
& \vec{b}=\langle-3,-7,6\rangle \\
& \vec{p}=\langle 3,-1,1\rangle \\
& \vec{q}=\langle-3,2,4\rangle \\
& \vec{p} \times \vec{q}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -1 & 1 \\
-3 & 2 & 4
\end{array}\right|=\langle-6,-15,3\rangle \\
& \text { Shortest distance }=\left|\frac{(\vec{b}-\vec{a}) \cdot(\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|}\right| \\
& =\left|\frac{(-6,-15,3) \cdot(-6,-15,3)}{\sqrt{36+225+9}}\right| \\
& =\left|\frac{36+225+9}{\sqrt{36+225+9}}\right| \\
& =\sqrt{270}=3 \sqrt{30} \\
&
\end{aligned}
\)

Hint

Plane passing through \((2,1,0),(4,1,1)\) and \((5,0,1)\) is
\(
\left|\begin{array}{lll}
x-2 & y-1 & z-0 \\
4-2 & 1-1 & 1-0 \\
5-2 & 0-1 & 1-0
\end{array}\right|=0
\)
\(\therefore\) Image of \(R(2,1,6)\) in this plane is
\(
\begin{aligned}
& \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}=-2 \frac{(2+1-12-3)}{1+1+4} \\
& \therefore( x , y , z )=(6,5,-2)
\end{aligned}
\)

Hint

Planes bisecting the given planes are
\(
\begin{aligned}
& \frac{2 x-y+2 z-4}{3}= \pm \frac{x+2 y+2 z-2}{3} \\
& \Rightarrow x -3 y =2 \text { or } 3 x + y +4 z =6
\end{aligned}
\)
Out of the four given points in question’s options only \((2,-4,1)\) lies on the plane \(3 x+y\) \(+4 z=6\)

Hint

Vector of the plane is
\(
\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -1 \\
-1 & 1 & -2
\end{array}\right|=-3 \hat{i}+3 \hat{j}+3 \hat{k}
\)
Now equation of plane is
\(
-3 x+3 y+3 z=c
\)
\((1,1,0)\) will satisfy the plane
\(
\begin{aligned}
& \Rightarrow-3+3+0=c \\
& \Rightarrow c =0 \\
& -3 x+3 y+3 z=0
\end{aligned}
\)
distance from \((2,1,4)\) is
\(
\Rightarrow\left|\frac{-6+3+12}{\sqrt{27}}\right|=\left|\frac{9}{3 \sqrt{3}}\right|=\sqrt{3} \text { units }
\)

Hint

\(
\begin{aligned}
& \frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}=\lambda \\
& A(3 \lambda+2,2 \lambda-1,-\lambda+1) \text { line on } 2 x+3 y-z+13=0 \\
& \Rightarrow 2(3 \lambda+2)+3(2 \lambda-1)-(-\lambda+1)+13=0 \\
& \Rightarrow 13 \lambda+13=0 \Rightarrow \lambda=-1
\end{aligned}
\)
Now point \(P(-1,-3,2)\) lie on \(3 x+y+4 z=16\)
\(
\begin{aligned}
& \Rightarrow 3(3 \lambda+2)+(3 \lambda+2)+4(3 \lambda+2)=16 \\
& \Rightarrow 9 \lambda+6+2 \lambda-4 \lambda-1+4=16 \\
& \Rightarrow 7 \lambda=7 \Rightarrow \lambda=1 \\
& \Rightarrow Q(5,1,0) \\
& \therefore P Q=\sqrt{36+16+4}=\sqrt{56}=2 \sqrt{14}
\end{aligned}
\)

Hint

\(
\text { (i) } \begin{aligned}
& \frac{|\lambda-6|}{\sqrt{16+4+16}}=\left|\frac{\lambda-6}{6}\right|=\frac{1}{3} \\
& \Rightarrow|\lambda-6|=2 \\
& \Rightarrow \lambda=8,4
\end{aligned}
\)
\(
\begin{aligned}
& \text { (ii) } \frac{|\mu-3|}{\sqrt{4+4+1}}=\frac{2}{3} \\
& \Rightarrow|\mu-3|=2 \\
& \Rightarrow \mu=5,1 \\
& \therefore(\mu+\lambda)_{\max }=13
\end{aligned}
\)

Hint

\(
\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}=\lambda
\)
Let a point \(P\) on the line is
\(
(2 \lambda+1,-\lambda-1,+\lambda)
\)
Foot of \(\perp^r Q\) is given by
\(
\frac{x-2 \lambda-1}{1}=\frac{y+\lambda+1}{1}=\frac{z-\lambda}{1}=-\frac{(2 \lambda-3)}{3}
\)
\(\therefore Q\) lies on \(x+y+z=3 \& x-y+z=3\)
\(
\begin{aligned}
& \Rightarrow x+z=3 \& y=0 \\
& \therefore y=0 \Rightarrow \lambda+1=\frac{-2 \lambda+3}{3} \Rightarrow \lambda=0
\end{aligned}
\)
\(\therefore Q\) is \((2,0,1)\)

Hint

\(
\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}=p P(\beta, 0, \beta)
\)
any point on line \(A=(p, 1,-p-1)\)
Now, DR of \(AP \equiv< p -\beta, 1-0,- p -1-\beta>\)
Which is perpendicular to line so
\(
\begin{aligned}
& ( p -\beta) \cdot 1+0.1-1(- p -1-\beta)=0 \\
& \Rightarrow p -\beta+ p +1+\beta=0 \\
& p=\frac{-1}{2}
\end{aligned}
\)
Point \(A\left(\frac{-1}{2}, 1-\frac{1}{2}\right)\)
Now, distance AP \(=\sqrt{\frac{3}{2}}\)
\(
\Rightarrow A P^2=\frac{3}{2}
[latex]
[latex]
\begin{aligned}
& \Rightarrow\left(\beta+\frac{1}{2}\right)^2+1+\left(\beta+\frac{1}{2}\right)^2=\frac{3}{2} \\
& \Rightarrow 2\left(\beta+\frac{1}{2}\right)^2=\frac{1}{2} \\
& \Rightarrow\left(\beta+\frac{1}{2}\right)^2=\frac{1}{4} \\
& \Rightarrow \beta=0,-1,(\beta \neq 0) \\
& \therefore \beta=-1
\end{aligned}
\)

Hint

Image of \(Q\) in plane
\(
\begin{aligned}
& \frac{(x-0)}{3}=\frac{(y+1)}{-1}=\frac{z+3}{+4}=\frac{-2(1-12-2)}{9+1+16}=1 \\
& x=3, y=-2, z=1 \\
& P(3,-2,1), Q(0,-1,-3), R(3,-1,-2)
\end{aligned}
\)
Now area of \(\triangle P Q R\) is
\(
\frac{1}{2}|\overline{P Q} \times \overline{Q R}|
\)
\(
=\frac{1}{2}\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
3 & -1 & 4 \\
3 & 0 & 1
\end{array}\right|
\)
\(
\begin{aligned}
& \Rightarrow \frac{1}{2}|\{\hat{i}(-1)-\hat{j}(3-12)+\widehat{k}(3)\}| \\
& \Rightarrow \frac{1}{2} \sqrt{(1+81+9)} \\
& \Rightarrow \frac{\sqrt{91}}{2}
\end{aligned}
\)

Hint

Let \(A D\) be a perpendicular drawn from point \(A\) on
The vertices \(B\) and \(C\) of a \(\triangle A B C\) lie on the line, \(\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}\) Coordinates of D can be expressed as \((3 \lambda-2,1,4 \lambda)\).
DR’s of AD are \((3 \lambda-3,2,4 \lambda-2)\)
\(A D\) is perpendicular to \(B C\)
\(
\begin{aligned}
& \therefore 3(3 \lambda-3)+0(2)+4(4 \lambda-2)=0 \\
& \Rightarrow 9 \lambda-9+0+16 \lambda-8=0 \\
& \Rightarrow \lambda=\frac{17}{25}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Coordinates of D are }\left(\frac{1}{25}, 1, \frac{68}{25}\right) \\
& \text { Length of } A D=\sqrt{\left(\frac{1}{25}-1\right)^2+(1-(-1))^2+\left(\frac{68}{25}-2\right)^2} \\
& =\sqrt{\frac{576}{625}+4+\frac{324}{625}} \\
& =\sqrt{\frac{3400}{625}}=\frac{2}{5} \sqrt{34} \\
&
\end{aligned}
\)
\(
\text { Area of triangle }=\frac{1}{2} \times 5 \times \frac{2}{5} \sqrt{34}=\sqrt{34} \text { sq. units }
\)

Hint

\(
\begin{aligned}
& P_1: x+y+z-6=0 \\
& P_2: 2 x+3 y+z+5=0
\end{aligned}
\)
Equation of plane which passes through the line of intersection of \(P_1\) and \(P_2\) is
\(
\begin{aligned}
& P_1+\lambda P_2=0 \\
& \Rightarrow(x+y+z-6)+\lambda(2 x+3 y+z+5)=0 \\
& \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1+\lambda)+(5 \lambda-6)=0
\end{aligned}
\)
As the above plane is perpendicular to xy plane
\(
\begin{aligned}
& \therefore \vec{n} \cdot \hat{k}=0 \\
& \Rightarrow[(2+\lambda) \hat{i}+(3+\lambda) \hat{j}+(1+\lambda) \widehat{k}] \cdot \widehat{k}=0 \\
& \Rightarrow 1+\lambda=0 \\
& \Rightarrow \lambda=-1
\end{aligned}
\)
So, equation of plane
\(
\begin{aligned}
& -x-2 y-11=0 \\
& \Rightarrow x+2 y+11=0
\end{aligned}
\)
Distance of the point \((0,0,256)\) from this plane
\(
=\left|\frac{0+0+11}{\sqrt{5}}\right|=\frac{11}{\sqrt{5}}
\)

Hint

Let \(a x+b y+c z=1\) be the equation of the plane
it passed through point \((0,-1,0)\).
\(
\begin{aligned}
& \therefore-b=1 \\
& \Rightarrow b=-1
\end{aligned}
\)
Also it passes through point \((0,0,1)\)
\(
\therefore c=1
\)
So the plane is \(a x-y+z=1\).
This plane an angle \(\frac{\pi}{4}\) with the plane \(y-z+5=0\).
Normal to the plane \(a x-y+z=1\) is
\(
\vec{a}=a \hat{i}-\hat{j}+\widehat{k}
\)
Normal to the plane \(y-z+5=0\) is
\(
\vec{b}=\hat{j}-\widehat{k}
\)
\(
\begin{aligned}
& \cos \theta=\left|\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right| \\
& \Rightarrow \frac{1}{\sqrt{2}}=\left|\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right| \\
& \Rightarrow \frac{|0-1-1|}{\sqrt{a^2+1+1} \sqrt{1^2+1^2}}=\frac{1}{\sqrt{2}} \\
& \Rightarrow a^2+2=4 \\
& \Rightarrow a= \pm \sqrt{2}
\end{aligned}
\)
\(\therefore\) Equation of plane
\(
\pm \sqrt{2} x-y+z=1
\)
Now by checking each options you can see equation \(-\sqrt{2} x-y+z=1\) satisfy by the point \((\sqrt{2}, 1,4)\)

Hint

Let \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}=\lambda\)
Any arbitary point on the line is \(P (2 \lambda+1,3 \lambda-1,4 \lambda+2)\).
This point also lies on the plane \(x+2 y+3 z=15\).
\(
\begin{aligned}
& \therefore(2 \lambda+1)+2(3 \lambda-1)+3(4 \lambda+2)=15 \\
& \Rightarrow 20 \lambda=10 \\
& \Rightarrow \lambda=\frac{1}{2}
\end{aligned}
\)
So point \(P\) is \(\left(2, \frac{1}{2}, 4\right)\).
Distance from the origin \((0)\) of point \(P\left(2, \frac{1}{2}, 4\right)\) is
\(
\begin{aligned}
& OP =\sqrt{(2)^2+\left(\frac{1}{2}\right)^2+(4)^2} \\
& =\sqrt{4+\frac{1}{4}+16} \\
& =\sqrt{\frac{81}{4}} \\
& =\frac{9}{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& P_1: x+y+z=1 \\
& P_1: 2 x+3 y+4 z=5
\end{aligned}
\)
Equation of the plane passing through the line of intersection of the plane \(P_1\) and \(P_2\) is
\(
\begin{aligned}
& P _1+\lambda P _2=0 \\
& \Rightarrow(x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 \\
& \Rightarrow x(1+2 \lambda)+y(1+3 \lambda)+z(1+4 \lambda)-5 \lambda-1=0 \dots(1)
\end{aligned}
\)
Direction Ratio (D.R) of this plane \(=(1+2 \lambda, 1+3 \lambda, 1+4 \lambda)\)
Plane (1) is perpendicular to \(x-y+z=0\), whose D.R \(=(1,-1,1)\)
As they are perpendicular so dot product of D.R \(=0\)
\(
\begin{aligned}
& \therefore(1)(1+2 \lambda)+(-1)(1+3 \lambda)+(1)(1+4 \lambda)=0 \\
& \Rightarrow 1+2 \lambda-1-3 \lambda+1+4 \lambda=0 \\
& \Rightarrow \lambda=-\frac{1}{3}
\end{aligned}
\)
Putting the value of \(\lambda\) in equation (1), we get
\(
\begin{aligned}
& \Rightarrow \frac{x}{3}-\frac{z}{3}+\frac{2}{3}=0 \\
& \Rightarrow x – z +2=0
\end{aligned}
\)
Vector form of this plane,
\(
\vec{r} \cdot(\hat{i}-\hat{k})+2=0
\)

Hint

Equation of \(P Q\) is
\(
\begin{aligned}
& \frac{x-2}{8-2}=\frac{y+3}{0-(-3)}=\frac{z-4}{10-4} \\
& \Rightarrow \frac{x-2}{6}=\frac{y+3}{3}=\frac{z-4}{6}
\end{aligned}
\)
Point \(R(4, y, z)\) lies on this
\(
\begin{aligned}
& \therefore \frac{4-2}{6}=\frac{y+3}{3}=\frac{z-4}{6} \\
& \Rightarrow \frac{1}{3}=\frac{y+3}{3}=\frac{z-4}{6}
\end{aligned}
\)
\(y=-2\) and \(y=6\)
\(
\therefore R=(4,-2,6)
\)
Distance of \(R(4,-2,6)\) from the origin \(O(0,0,0)\) is
\(
\begin{aligned}
& RO =\sqrt{4^2+(-2)^2+6^2} \\
& =\sqrt{56}=2 \sqrt{14}
\end{aligned}
\)

Hint

Given,
\(P=(2,-1,4)\) and equation of line \(\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}\)
Any point \(B\) on the line is \((10 \lambda-3,-7 \lambda+2, \lambda)\)
Now,
Direction ratios of \(PB =(10 \lambda-5,-7 \lambda+3, \lambda-4)\)
Direction ratios of the line \(=(10,-7,1)\)
As the lines are perpendicular, therefore dot product of given lines are 0 .
\(
\begin{aligned}
& \Rightarrow 10(10 \lambda-5)-7(-7 \lambda+3)+(\lambda-4)=0 \\
& \Rightarrow 100 \lambda-50+49 \lambda-21+\lambda-4=0 \\
& \Rightarrow 150 \lambda=75 \\
& \Rightarrow \lambda=\frac{1}{2}
\end{aligned}
\)
Point \(B=\left(2,-\frac{3}{2}, \frac{1}{2}\right)\)
\(
|P B|=\frac{5}{\sqrt{2}}=3.53
\)

Hint

Let vector \(\vec{p}\) is perpendicular to the both vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{i}+\hat{2 j}+\hat{k}\).
\(
\begin{aligned}
& \therefore \vec{p}=(\hat{i}+\hat{j}+\hat{k}) \times(\hat{i}+\hat{2 j}+3 \hat{k}) \\
& =\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
1 & 2 & 3
\end{array}\right| \\
& =\hat{i}-2 \hat{j}+\widehat{k}
\end{aligned}
\)
Now a vector \(\vec{a}=\hat{2 i}+\hat{3 j}+\hat{k}\) is given and we have to findout projection of vector \(\vec{a}\) on \(\vec{p}\).
\(\therefore\) Projection of vector \(\vec{a}\) on \(\vec{p}\)
\(
\begin{aligned}
& =|\vec{a}| \cos \theta \\
& =|\vec{a}| \times \frac{\overrightarrow{a \cdot p}}{|\vec{a}||\vec{p}|} \\
& =\frac{\vec{a} \cdot \vec{p}}{|\vec{p}|} \\
& =\frac{(2 \hat{i}+3 \hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+\hat{k})}{\sqrt{1+4+1}} \\
& =\frac{2-6+1}{\sqrt{6}} \\
& =\frac{-3}{\sqrt{6}}
\end{aligned}
\)
Magnitude of projection of vector \(\vec{a}\) on \(\vec{p}\)
\(
=\left|\frac{-3}{\sqrt{6}}\right|=\frac{3}{\sqrt{6}}=\frac{\sqrt{3}}{\sqrt{2}}
\)

Hint

The equation of any plane passing through the intersection of the planes \(2 x-y-4=\) 0 and \(y+2 z-4=0\) is :
\(
(2 x-y-4)+\lambda(y+2 z-4)=0 \dots(1)
\)
As this plane passes through \((1,1,0)\) then this point satisfy the equation (1).
\(
\begin{aligned}
& \therefore(2-1-4)+\lambda(1+0-4)=0 \\
& \Rightarrow \lambda=-1
\end{aligned}
\)
Equation of required plane will be
\(
\begin{aligned}
& (2 x-y-4)-(y+2 z-4)=0 \\
& \Rightarrow 2 x-2 y-2 z=0 \\
& \Rightarrow x-y-z=0
\end{aligned}
\)

Hint

All four points are coplanar so
\(
\begin{aligned}
& \left|\begin{array}{ccc}
1-\lambda^2 & 2 & 0 \\
2 & -\lambda^2+1 & 0 \\
2 & 2 & -\lambda^2-1
\end{array}\right|=0 \\
& \left(\lambda^2+1\right)^2\left(3-\lambda^2\right)=0 \\
& \lambda= \pm \sqrt{3}
\end{aligned}
\)

Hint

DR’s of line are \(2,1,-2\)
normal vector of plane is \(\hat{i}-2 \hat{j}- k \widehat{k}\)
\(
\sin \alpha=\frac{(2 \hat{i}+\hat{j}-2 \hat{k}) \cdot(\hat{i}-2 \hat{j}-k \hat{k})}{3 \sqrt{1+4+k^2}}
\)
\(
\begin{aligned}
&\begin{aligned}
& \sin \alpha=\frac{2 k}{3 \sqrt{k^2+5}} \quad \ldots \ldots . (1)\\
& \cos \alpha=\frac{2 \sqrt{2}}{3} \quad \ldots \ldots .(2)
\end{aligned}\\
&(1)^2+(2)^2=1 \Rightarrow k^2=\frac{5}{3}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \left|\begin{array}{lll}
i & j & k \\
3 & 5 & 7 \\
1 & 4 & 7
\end{array}\right| \\
& =\hat{i}(35-28)-\hat{j}(21.7)+\widehat{k}(12-5)
\end{aligned}
\)
\(
\begin{aligned}
& =7 \hat{i}-14 \hat{j}+7 \widehat{k} \\
& =\hat{i}-2 \hat{j}+\widehat{k} \\
& 1( x +2)-2( y -2)+1( z +15)=0 \\
& x -2 y + z +11=0 \\
& \frac{11}{\sqrt{4+1+1}}=\frac{11}{\sqrt{6}}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \overrightarrow{O P} \times \overrightarrow{O Q}=(\hat{i}+2 \hat{j}+\widehat{k}) \times(2 \hat{i}+\hat{j}+3 \widehat{k}) \\
& =5 \hat{i}-\hat{j}-3 \widehat{k} \\
& \overrightarrow{P Q} \times \overrightarrow{P R}=(\hat{i}-\hat{j}+2 \widehat{k}) \times(-2 \hat{i}-\hat{j}+\widehat{k}) \\
& =\hat{i}-5 \hat{j}-3 \widehat{k} \\
& \cos \theta=\frac{5+5+9}{(\sqrt{25+9+1})^2}=\frac{19}{35}
\end{aligned}
\)

Hint

Point on \(L _1(\lambda+3,3 \lambda-1,-\lambda+6)\)

Point on \(L _2(7 \mu-5,-6 \mu+2,4 \mu+3\)
\(
\Rightarrow \lambda+3=7 \mu-5
\)
\(3 \lambda-1=-6 \mu+2 \dots(i)\)
\(\Rightarrow \lambda=-1, \mu=1 \dots(ii)\)
point \(R(2,-4,7)\)
Reflection is \((2,-4,-7)\)

Hint

Normal vector of plane
\(
\begin{aligned}
& =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
2 & -5 & 0 \\
4 & -4 & 5
\end{array}\right| \\
& =-4(5 \hat{i}+2 \hat{j}-3 \widehat{k})
\end{aligned}
\)
equation of plane is
\(
\begin{aligned}
& 5(x-7)+2 y-3(z-6)=0 \\
& 5 x+2 y-3 z=17 \\
& 5 \times 2+2 \alpha-3 \beta=17 \\
& \therefore 2 \alpha-3 \beta=17-10=7
\end{aligned}
\)

Hint

The normal vector of required plane
\(
\begin{aligned}
& =(2 \hat{i}-\hat{j}+3 \widehat{k}) \times(2 \hat{i}+3 \hat{j}-\widehat{k}) \\
& =-8 \hat{i}+8 \hat{j}+8 \widehat{k}
\end{aligned}
\)
So, direction ratio of normal is \((-1,1,1)\)
So required plane is
\(
\begin{aligned}
& -(x-3)+(y+2)+(z-1)=0 \\
& \Rightarrow-x+y+z+4=0
\end{aligned}
\)
Which is satisfied by \((2,0,-2)\)

Hint

Let the equation of plane be
\(
a(x-0)+b(y+1)+c(z-0)=0
\)
It passes through \((0,0,1)\) then
\(
b+c=0 \quad \ldots .(1)
\)
Now \(\cos \frac{\pi}{4}=\frac{a(0)+b(1)+c(-1)}{\sqrt{2} \sqrt{a^2+b^2+c^2}}\)
\(\Rightarrow a ^2=-2 bc\) and \(b =- c\)
we get \(a^2=2 c^2\)
\(
\Rightarrow a = \pm \sqrt{2} c
\)
\(\Rightarrow\) direction ratio \(( a , b , c )=(\sqrt{2},-1,1)\) or \((\sqrt{2}, 1,-1)\)

Hint

General point on the given line is
\(
\begin{aligned}
& x=2 \lambda+4 \\
& y=2 \lambda+5 \\
& z=\lambda+3
\end{aligned}
\)
Solving with plane,
\(
\begin{aligned}
& 2 \lambda+4+2 \lambda+5+\lambda+3=2 \\
& 5 \lambda+12=2 \\
& 5 \lambda=-10 \\
& \lambda=-2
\end{aligned}
\)

Hint

Since, direction ratios of normal to the plane is
\(
\vec{n}=6 \hat{i}+10 \hat{j}+2 \hat{k}
\)
Then, equation of the plane is
\(
\begin{aligned}
& (x-0) 6+(y-2) 10+(z-5) 2=0 \\
& 3 x+5 y-10+z-5=0 \\
& 3 x+5 y+z=15 \dots(1)
\end{aligned}
\)
Since, plane (1) satisfies the point \((4,1,-2)\)
Hence, required point is \((4,1,-2)\)

Hint

Equation of required plane is
\(
\left|\begin{array}{ccc}
x-4 & y+1 & z-2 \\
3 & -1 & 2 \\
1 & 2 & 3
\end{array}\right|=0
\)
\(
(x-4)(-3-4)-\left(y+1\right)(9-2)+(z-2)(6+1)=0
\)
\(
\begin{aligned}
& \Rightarrow-7(x-4)-7(y+1)+7(z-2)=0 \\
& \Rightarrow x-4+y+1-z+2=0 \\
& \Rightarrow x+y-z-1=0
\end{aligned}
\)
\(\because\) point \((1,1,1)\) satisfies this equation
\(\therefore\) point \((1,1,1)\) lies on the plane

Hint

\(\because A\) be a point on given line.
\(\therefore\) Position vector of \(A\)
\(
\overrightarrow{O A}=(1-3 \mu) \hat{i}+(\mu-1) \hat{j}+(2+5 \mu) \hat{k}
\)
Position vector of \(B=\overrightarrow{O B}=3 \hat{i}+2 \hat{j}+6 \hat{k}\)
\(
\begin{aligned}
& \therefore \quad \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\
&=(3 \mu+2) \hat{i}+(3-\mu) \hat{j}+(4-5 \mu) \hat{k}
\end{aligned}
\)
Equation of plane is: \(x-4 y+3 z=1\)
\(\because \overrightarrow{A B}\) is parallel to this plane.
\(
\begin{aligned}
& \therefore 1(3 \mu+2)-4(3-\mu)+3(4-5 \mu)=0 \\
& \Rightarrow 3 \mu+2-12+4 \mu+12-15 \mu=0 \\
& \Rightarrow 2-8 \mu=0 \\
& \Rightarrow \mu=\frac{1}{4}
\end{aligned}
\)

Hint

\(
\text { Vector } \perp \text { to given plane }=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 4 & 2 \\
4 & 2 & 3
\end{array}\right|
\)
\(
\begin{aligned}
& =\hat{i}(12-4)-\hat{j}(9-8)+\widehat{k}(6-16) \\
& =8 \hat{i}-\hat{j}-10 \widehat{k} \quad \ldots(1)
\end{aligned}
\)
Vector parallel to given line
\(
=2 \hat{i}+3 \hat{j}+4 \widehat{k} \dots(2)
\)
Vector \(\perp\) to both (1) & (2) vector
\(
\begin{aligned}
& =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
8 & -1 & -10 \\
2 & 3 & 4
\end{array}\right| \\
& =\hat{i}(-4+30)-\hat{j}(32+20)+\widehat{k}(24+2) \\
& =26 \hat{i}-52 \hat{j}+26 \widehat{k}
\end{aligned}
\)
Dr’s of normal of required plane is
\(
(26,-52,26) \Rightarrow(1,-2,1)
\)
Equation of plane whose Dr’s of Normal is \((1,-2,1)\) and passes through origin
\(
\begin{aligned}
& \text { 1. }(x-0)-2(y-0)+1 \cdot(z-0)=0 \\
& x-2 y+z=0
\end{aligned}
\)

Hint

Equation of \(1^{\text {st }}\) line is
\(
\frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}
\)
Dr’s of \(1^{\text {st }}\) line \(=(a, 1, c)\)
Equation of \(2^{\text {nd }}\) line is
\(
\frac{x-b^{\prime}}{a^{\prime}}=\frac{y-b^{\prime}}{c^{\prime}}=\frac{z}{1}
\)
Dr’s of \(2^{\text {nd }}\) line \(=\left(a^{\prime}, c^{\prime}, 1\right)\)
Lines are perpendicular, so the dot product of the Dr’s of two lines are zero.
\(
\therefore a a^{\prime}+c+c^{\prime}=0
\)

Hint

Let any point on the intersecting line
\(
\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1}=\lambda \text { (say) }
\)
is \(\quad(-3 \lambda-1,2 \lambda+3,-\lambda+2)\)
Since, the above point lies on a line which passes through the point \((-4,3,1)\)
Then, direction ratio of the required line
\(
\begin{aligned}
& =<-3 \lambda-1+4,2 \lambda+3-3,-\lambda+2-1\rangle> \\
& \text { or }\langle-3 \lambda+3,2 \lambda,-\lambda+1\rangle
\end{aligned}
\)
Since, line is parallel to the plane
\(
x+2 y-z-5=0
\)
Then, perpendicular vector to the line is \(\hat{i}+2 \hat{j}-\hat{k}\)
Now \((-3 \lambda+3)(1)+(2 \lambda)(2)+(-\lambda+1)(-1)=0\)
\(
\Rightarrow \lambda=-1
\)
Now direction ratio of the required line \(=\langle 6,-2,2\rangle\) or \(<3,-1,1>\)
Hence required equation of the line is
\(
\frac{(x+4)}{3}=\frac{y-3}{-1}=\frac{z-1}{1}
\)

Hint

The equation of plane
\(
\begin{aligned}
& (x+y+z-1)+\lambda(2 x+3 y-z+4)=0 \\
& \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+4 \lambda-1=0
\end{aligned}
\)
As plane is parallel to \(y\) axis so the normal vector of plane and dot product of \(\hat{j}\) is zero.
\(
\begin{aligned}
& \therefore 1+3 \lambda=0 \\
& \Rightarrow \lambda=-\frac{1}{3}
\end{aligned}
\)
\(\therefore\) So the equation of the plane is
\(
\begin{aligned}
& x\left(1-\frac{2}{3}\right)+\left(1-\frac{3}{3}\right) y+\left(1+\frac{1}{3}\right)-\frac{4}{3}-1=0 \\
& \Rightarrow x\left(\frac{1}{3}\right)+z\left(\frac{4}{3}\right)-\frac{7}{3}=0 \\
& \Rightarrow x+4 z-7=0
\end{aligned}
\)
By checking each options you can see only point \((3,2,1)\) lies on the plane.

Hint

Let \(\theta\) be the angle between the two lines
Here direction cosines of \(\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\) are \(2,2,1\)
Also second line can be written as :
\(
\frac{x-5}{2}=\frac{y-2}{\frac{p}{7}}=\frac{z-3}{4}
\)
\(\therefore\) its direction cosines are \(2, \frac{P}{7}, 4\)
Also, \(\cos \theta=\frac{2}{3}\) (Given)
\(
\begin{aligned}
& \because \cos \theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2 \sqrt{a_2^2+b_2^2+c_2^2}}}\right| \\
& \Rightarrow \quad \frac{2}{3}=\left|\frac{(2 \times 2)+\left(2 \times \frac{P}{7}\right)+(1 \times 4)}{\sqrt{2^2+2^2+1^2} \sqrt{2^2+\frac{P^2}{49}+4^2}}\right| \\
& =\frac{4+\frac{2 P}{7}+4}{3 \times \sqrt{2^2+\frac{P^2}{49}+4^2}} \\
& \Rightarrow\left(4+\frac{P}{7}\right)^2=20+\frac{P^2}{49} \\
& \Rightarrow 16+\frac{8 P}{7}+\frac{P^2}{49}=20+\frac{P^2}{49} \\
& \Rightarrow \frac{8 P}{7}=4 \Rightarrow P=\frac{7}{2} \\
&
\end{aligned}
\)

Hint

Equation of plane passing through three given points is :
\(
\begin{aligned}
& \left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
x+2 & y+2 & z-2 \\
1+2 & -1+2 & 2-2 \\
1+2 & 1+2 & 1-2
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
x+2 & y+2 & z-2 \\
3 & 1 & 0 \\
3 & 30 & -1
\end{array}\right|=0 \\
& \Rightarrow-x+3 y+6 z-8=0 \\
& \Rightarrow \frac{x}{8}-\frac{3 y}{8}-\frac{6 z}{8}+\frac{8}{8}=0 \\
& \Rightarrow \frac{x}{8}-\frac{y}{\frac{8}{3}}-\frac{z}{\frac{8}{6}}=-1 \\
& \Rightarrow \frac{x}{-8}+\frac{y}{\frac{8}{3}}+\frac{z}{\frac{8}{6}}=1
\end{aligned}
\)
\(
\therefore \text { Sum of intercepts }=-8+\frac{8}{3}+\frac{8}{6}=-4
\)

Hint

Plane passing through line of intersection of first two planes is
\(
(2 x-2 y+3 z-2)+\lambda(x-y+z+1)=0
\)
\(x(\lambda+2)-y(2+\lambda)+z(\lambda+3)+(\lambda-2)=0 \dots(i)\)
is having infinite number of solutions with \(x+2 y-z-3=0\) and \(3 x-y+2 z-1=0\) then
\(
\left|\begin{array}{ccc}
\lambda+2 & -(\lambda+2) & \lambda+3 \\
1 & 2 & -1 \\
3 & -1 & 2
\end{array}\right|=0
\)
On solving, we have \(\lambda=5\)
Hence required equation of plane is \(7 x-7 y+8 z+3=0\)
Now, perpendicular distance from \((0,0,0)\) is \(=\frac{3}{\sqrt{49+49+64}}\)
\(=\frac{3}{\sqrt{162}}=\frac{1}{3 \sqrt{2}}\)

Hint

\(P Q\) is the projection of line segment \(A B\) on the plane \(x+y+z=7\)
\(P\) and \(Q\) are called foot of perpendicular on the plane \(x+y+z=7\)
Let \(P=(x, y, z)\) then
\(
\begin{aligned}
& \frac{x-5}{1}=\frac{y+1}{1}=\frac{z-4}{1}=\frac{-(5-1+4)}{1^2+1^2+1^2} \\
& \Rightarrow x-5=y+1=z-4=-\frac{8}{3} \\
& \therefore x =\frac{7}{3}, y =-\frac{11}{3}, z =\frac{4}{3} \\
& \therefore \text { Point } P =\left(\frac{7}{3},-\frac{11}{3}, \frac{4}{3}\right)
\end{aligned}
\)
Let \(Q =\left( x _1, y _1, z _1\right)\) then
\(
\begin{aligned}
& \frac{a_1-4}{1}=\frac{y_1+1}{1}=\frac{z_1-3}{1}=\frac{-(4-1+3)}{1^2+1^2+1^2} \\
& \Rightarrow x _1-4= y _1+1= z _1-3=-2 \\
& \therefore x =2, y =-3, z =1 \\
& \therefore \text { Point } Q =(2,-3,1)
\end{aligned}
\)
Now length of \(PQ[latex] is
[latex]
\sqrt{\left(\frac{7}{3}-2\right)^2+\left(-\frac{11}{3}+3\right)^2+\left(\frac{4}{3}-1\right)^2}=\sqrt{\frac{2}{3}}
\)

Hint

Given \(l+3 m+5 n=0 \dots(1)\)
and \(5 l m-2 m n+6 n l=0 \dots(2)\)
From eq. (1) we have
\(
l=-3 m-5 n
\)
Put the value of \(l\) in eq. (2), we get ;
\(
\begin{aligned}
& 5(-3 m-5 n) m-2 m n+6 n(-3 m-5 n)=0 \\
& \Rightarrow 15 m^2+45 m n+30 n^2=0 \\
& \Rightarrow m^2+3 m n+2 n^2=0 \\
& \Rightarrow m^2+2 m n+m n+2 n^2=0 \\
& \Rightarrow(m+n)(m+2 n)=0 \\
& \therefore m=-n \text { or } m=-2 n
\end{aligned}
\)
For \(m =-n, l=-2 n\)
And for \(m =-2 n , l= n\)
\(
\begin{aligned}
& \therefore(l, m , n )=(-2 n ,- n , n ) \operatorname{Or}(l, m , n )=( n ,-2 n , n ) \\
& \Rightarrow(l, m , n )=(-2,-1,1) \operatorname{Or}(l, m , n )=(1,-2,1)
\end{aligned}
\)
Therefore, angle between the lines is given as :
\(
\begin{aligned}
& \cos (\theta)=\frac{(-2)(1)+(-1) \cdot(-2)+(1)(1)}{\sqrt{6} \cdot \sqrt{6}} \\
& \Rightarrow \cos (\theta)=\frac{1}{6} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{6}\right)
\end{aligned}
\)

Hint

Since the plane bisects the line joining the points \((1,2,3)\) and \((-3,4,5)\) then the plane passes through the midpoint of the line which is:
\(
\left(\frac{1-3}{2}, \frac{2+4}{2}, \frac{5+3}{2}\right) \equiv\left(\frac{-2}{2}, \frac{6}{2}, \frac{8}{2}\right) \equiv(-1,3,4) \text {. }
\)
As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are \((-3-1,4-2,5-3)=(-4,2,2)\)
So the equation of the plane is : \(-4 x+2 y+2 z=\lambda\)
As plane passes through \((-1,3,4)\)
So, \(-4(-1)+2(3)+2(4)=\lambda \Rightarrow \lambda=18\)
Therefore, equation of plane is : \(-4 x+2 y+2 z=18\)
Now, only \((-3,2,1)\) satiesfies the given plane as
\(
-4(-3)+2(2)+2(1)=18
\)

Hint

If \(a, b, c\) are the intercepts of the variable plane on the \(x, y, z\) axes respectively, then the equation of the plane is
\(
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
\)
And the point of intersection of the planes parallel to the \(x y, y z\) and \(z x\) planes is \((a, b, c)\)
As the point \((3,2,1)\) lies on the variables plane,
\(
\text { so } \frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1
\)
Therefore, the required locus is
\(
\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1
\)

Hint

Normal to \(3 x+4 y+z=1\) is \(3 \hat{i}+4 \hat{j}+\widehat{k}\)
Normal to \(5 x+8 y+2 z=-14\) is \(5 \hat{i}+8 \hat{j}+2 \widehat{k}\)
The line of intersection of the planes is perpendicular to both normals, so, direction ratios of the intersection line are directly proportional to the cross product of the normal vectors.
Therefore the direction ratios of the line is \(-\hat{j}+4 \widehat{k}\)
Hence the angle between the plane \(x+y+z+5=0\)
and the intersection line is \(\sin ^{-1}\left(\frac{-1+4}{\sqrt{17} \sqrt{3}}\right)=\sin ^{-1}\left(\sqrt{\frac{3}{17}}\right)\)

Hint

Let \(M(3 \lambda+1,2 \lambda, 4 \lambda+2)\)
\(
\begin{aligned}
& \overrightarrow{ AM } \cdot \overrightarrow{ b }=0 \\
& \Rightarrow \quad 9 \lambda-15+4 \lambda-2+16 \lambda-12=0 \\
& \Rightarrow \quad 29 \lambda=29 \\
& \Rightarrow \quad \lambda=1 \\
& M (4,2,6), I =(2,3,7)
\end{aligned}
\)
\(
\text { Required Distance }=\sqrt{4+9+49}=\sqrt{62}
\)

Hint

\(
\begin{aligned}
& \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} \\
& \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k} \\
& A(t, 2 t+1,3 t+2) \\
& \overrightarrow{Q A}=(t-1) i+(2 t-5) j+(3 t-2) \hat{k} \\
& \overrightarrow{Q A} \cdot \vec{b}=0 \\
& (t-1)+2(2 t-5)+3(3 t-2)=0 \\
& 14 t=17 \\
& \alpha=\frac{20}{14} \\
& \beta=\frac{12}{14} \\
& \gamma=\frac{102}{14} \\
& 2 \alpha+\beta+\gamma=\frac{154}{14}=11 \text { Ans. }
\end{aligned}
\)

Hint

\(
\begin{aligned}
& L_1: \frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1} \\
& L_2: \frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4} \\
& n_1 \times n_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -1 & 1 \\
-3 & 2 & 4
\end{array}\right| \\
& =-6 \hat{i}-15 \hat{j}+3 \hat{k}
\end{aligned}
\)
\(
\begin{aligned}
& d=\left|\frac{[(\lambda+2) \hat{i}+7 \hat{j}-3 \hat{k}][-6 \hat{i}-15 \hat{j}+3 \hat{k}]}{|-6 \hat{i}-15 \hat{j}+3 \hat{k}|}\right|=\frac{44}{\sqrt{30}} \\
& \left|\frac{-6 \lambda-12-105-9}{\sqrt{270}}\right|=\frac{44}{\sqrt{30}} \\
& |6 \lambda+126|=132 \\
& |\lambda+21|=22 \\
& \lambda+21= \pm 22 \\
& |\lambda|_{\max }=43
\end{aligned}
\)

Hint

\(
\begin{aligned}
& P(10,-2,-1) \\
& M N: \frac{x-2}{8}=\frac{y+5}{-12}=\frac{z-11}{16} \\
& \text { General point } \\
& (8 k+2,-12 k-5,16 k+11) \\
& \overrightarrow{R Q}=(8 k+2-1) \hat{i}+(-12 k-5-7) \hat{j}+(16 k+11-6) \widehat{k} \\
& \overrightarrow{R Q}=(8 k+1) \hat{i}-(12 k+12) \hat{j}+(16 k+5) \widehat{k} \\
& \overrightarrow{R Q} \cdot \overrightarrow{M N}=0(\text { as both are perpendicular }) \\
& 8(8 k+1)+12(12 k+12)+16(16 k+5)=0 \\
& 64 k+8+144 k+144+256 k+80=0 \\
& 464 k=-232 \\
& k=\frac{-232}{464}=\frac{-1}{2} \\
& Q(-4+2,6-5,-8+11) \\
& Q(-2,1,3) \\
& P Q=\sqrt{(10+2)^2+(-3)^2+(4)^2} \\
& P Q=\sqrt{12^2+3^2+4^2} \\
& P Q=\sqrt{169}=13
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \overrightarrow{A B} \perp \vec{L}_1 \text { and } \overrightarrow{A B} \perp \vec{L}_2 \\
& \overrightarrow{A B}=(-3 m-2+n+2,4 m+2-2 n+6,2 m+5-1) \\
& =(-3 m+n, 4 m-2 n+8,2 m+4) \\
& \overrightarrow{A B} \perp \vec{L}_1 \\
& \Rightarrow-3(-3 m+n)+4(4 m-2 n+8)+2(2 m+4)=0 \\
& (9 m+16 m+4 m)+(-3 n-8 n)+32+8=0 \\
& \Rightarrow 29 m-11 n+40=0 \quad \ldots(1) \\
& \overrightarrow{A B} \perp \vec{L}_2 \\
& \Rightarrow-1(-3 m+n)+2(4 m-2 n+8)+0(2 m+4)=0 \\
& \Rightarrow 3 m-n+8 m-4 n+16=0 \\
& \Rightarrow 11 m-5 n+16=0 \Rightarrow m=-1, n=1 \\
& \Rightarrow A \equiv(1,-2,3), \quad B \equiv(-3,-4,1) \\
& A B \text { line } \Rightarrow \frac{x-1}{2}=\frac{y+2}{1}=\frac{z-3}{1} \\
& \Rightarrow \alpha=-2, \quad \beta=3 \\
& \Rightarrow(\alpha-\beta)^2=25
\end{aligned}
\)

Hint

DR’s of Line \(L \equiv-1: 1: 2\)
DR’s of \(A B \equiv \alpha-2: \beta-2: \gamma-2\)
\(
\begin{aligned}
& AB \perp L \Rightarrow 2-\alpha+\beta-2+2 \gamma-4=0 \\
& 2 \gamma+\beta-\alpha=4 \quad \ldots \text { (1) }
\end{aligned}
\)
Let \(C\) is mid-point of \(AB\)
\(
C \left(\frac{\alpha+2}{2}, \frac{\beta+2}{2}, \frac{\gamma+2}{2}\right)
\)
DR’s of PC \(=\frac{\alpha}{2}: \frac{\beta-2}{2}: \frac{\gamma}{2}\)
line \(L \| PC \Rightarrow \frac{-\alpha}{2}=\frac{\beta-2}{2}=\frac{\gamma}{4}= K\) (let)
\(
\begin{aligned}
& \alpha=-2 K \\
& \beta=2 K +2 \\
& \gamma=4 K
\end{aligned}
\)
Use in (1) \(\Rightarrow K=\frac{1}{6}\)
Value of \(\alpha+\beta+6 \gamma=24 K+2=6\)

Hint

\(
\begin{array}{ll}
L_1 \equiv \vec{r}=(1,2,3)+\lambda(1,-1,1) & \left(\vec{r}=\vec{a}_1+\lambda \vec{b}_1\right) \\
L_2 \equiv \vec{r}=(4,5,6)+\mu(1,1,-1) & \left(\vec{r}=\vec{a}_2+\lambda \vec{b}_2\right) \\
P \equiv(\lambda+1,-\lambda+2, \lambda+3) & \\
Q \equiv(\mu+4, \mu+5,-\mu+6)
\end{array}
\)
\(
\overrightarrow{P Q}=(\mu-\lambda+3, \mu+\lambda+3,-\mu-\lambda+3)
\)
\(
\overrightarrow{P Q} \cdot \vec{b}_1=0 \Rightarrow 3 \lambda+\mu=3 \dots(i)
\)
\(
\overrightarrow{P Q} \cdot \vec{b}_2=0 \Rightarrow 3 \mu+\lambda=3 \dots(ii)
\)
From (i) and (ii),
\(
\begin{aligned}
& P \equiv\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right) \& Q \equiv\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right) \\
& \alpha=\frac{5}{2}, \beta=\frac{4}{2}, \gamma=\frac{12}{2} \\
& 2(\alpha+\beta+\gamma)=21
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6} \\
& \frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21 \\
& \left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right) \\
& =(22,0,7)=(a, b, c) \\
& \therefore \sqrt{324+144+16}=22
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r \rightarrow Q(r, r, 1) \\
& \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1)
\end{aligned}
\)
\(
\begin{aligned}
& \overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k} \\
& a=r+a-r=0 \\
& 2 a=2 r \rightarrow a=r \\
& \overline{P R}=(a-k) i+(a+k) \hat{j}+(a+1) \hat{k} \\
& a-k-a-k=0 \Rightarrow k=0
\end{aligned}
\)
\(
\begin{aligned}
& A s, P Q \perp P R \\
& (a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0 \\
& a=1 \text { or }-1 \\
& 12 a^2=12
\end{aligned}
\)

Hint

\(
\begin{aligned}
& M (3 \lambda+1,2 \lambda+2,-2 \lambda-1) \quad \therefore \alpha+\beta+\gamma=3 \lambda+2 \\
& N (-3 \mu-2,-2 \mu+2,4 \mu+1) \quad \therefore a + b + c =-\mu+1
\end{aligned}
\)
\(
\begin{aligned}
& \frac{3 \lambda+2}{-3 \mu-1}=\frac{2 \lambda}{-2 \mu}=\frac{-2 \lambda-4}{4 \mu-2} \\
& 3 \lambda \mu+2 \mu=3 \lambda \mu+\lambda \\
& 2 \mu=\lambda \\
& 2 \lambda \mu-\lambda=\lambda \mu+2 \mu \\
& \lambda \mu=\lambda+2 \mu \\
& \Rightarrow \lambda \mu=2 \lambda \\
& \Rightarrow \quad \mu=2 \quad(\lambda \neq 0) \\
& \therefore \quad \lambda=4 \\
& \quad \alpha+\beta+\gamma=14 \\
& \quad a+b+c=-1 \\
& \frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}=196
\end{aligned}
\)

Hint

\(
\begin{aligned}
& L _1: \frac{ x +1}{1}=\frac{ y }{1 / 2}=\frac{ z }{-1 / 12}, L _2: \frac{ x }{1}=\frac{ y +2}{1}=\frac{ z -1}{\frac{1}{6}} \\
& d _1=\text { shortest distance between } L _1 \& L _2 \\
& =\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}\right| \\
& d_1=2 \\
& L _3: \frac{ x -1}{2}=\frac{ y +8}{-7}=\frac{ z -4}{5}, L _4: \frac{ x -1}{2}=\frac{ y -2}{1}=\frac{ z -6}{-3} \\
& d _2=\text { shortest distance between } L _3 \& L _4 \\
& d _2=\frac{12}{\sqrt{3}} \text { Hence } \\
& =\frac{32 \sqrt{3} d _1}{ d _2}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}=16
\end{aligned}
\)

Hint

\(
\begin{aligned}
& L _1: \frac{ x -5}{4}=\frac{ y -4}{1}=\frac{ z -5}{3}=\lambda \quad \operatorname{drs}(4,1,3)= b _1 \\
& M (4 \lambda+5, \lambda+4,3 \lambda+5) \\
& L _2: \frac{ x +8}{12}=\frac{ y +2}{5}=\frac{ z +11}{9}=\mu \\
& N (12 \mu-8,5 \mu-2,9 \mu-11) \\
& \overrightarrow{ MN }=(4 \lambda-12 \mu+13, \lambda-5 \mu+6,3 \lambda-9 \mu+16) \dots(1)
\end{aligned}
\)
\(
\overrightarrow{ b }_1 \times \overrightarrow{ b }_2=\left|\begin{array}{ccc}
\hat{ i } & \hat{ j } & \hat{ k } \\
4 & 1 & 3 \\
12 & 5 & 9
\end{array}\right|=-6 \hat{ i }+8 \hat{ k } \dots(2)
\)
Equation (1) and (2)
\(
\therefore \frac{4 \lambda-12 \mu+13}{-6}=\frac{\lambda-5 \mu+6}{0}=\frac{3 \lambda-9 \mu+16}{8}
\)
Eqn 1 and 2 gives
\(
\lambda-5 \mu+6=0 \dots(3)
\)
Eqn 1 and 3 gives
\(
\lambda-3 \mu+4=0 \dots(4)
\)
Solve (3) and (4) we get
\(
\begin{aligned}
\lambda & =-1, \mu=1 \\
\therefore & M (1,3,2) \\
& N (4,3,-2) \\
\therefore & \overrightarrow{ OM } \cdot \overrightarrow{ ON }=4+9-4=9
\end{aligned}
\)

Hint

Let \(P ( t , t -2, t )\) and \(Q (2 s -2, s , s )\)
D.R’s of \(P Q\) are \(2,1,2\)
\(
\begin{aligned}
& \frac{2 s -2- t }{2}=\frac{ s – t +2}{1}=\frac{ s – t }{2} \\
& \Rightarrow t =6 \text { and } s =2 \\
& \Rightarrow P (6,4,6) \text { and } Q (2,2,2) \\
& PQ : \frac{ x -2}{2}=\frac{ y -2}{1}=\frac{ z -2}{2}=\lambda
\end{aligned}
\)
Let \(F (2 \lambda+2, \lambda+2,2 \lambda+2)\)
\(
A (1,2,12)
\)
\(
\overrightarrow{ AF } \cdot \overrightarrow{ PQ }=0
\)
\(
\therefore \lambda=2
\)
So \(F(6,4,6)\) and \(A F=\sqrt{65}\)

Hint

\(
\begin{aligned}
& \frac{ x -2}{1}=\frac{ y }{-1}=\frac{ z -7}{8}=\lambda \\
& \frac{ x +3}{4}=\frac{ y +2}{3}=\frac{ z +2}{1}= k \\
& \Rightarrow \lambda+2=4 k -3 \\
& -\lambda=3 k -2 \\
& \Rightarrow k =1, \lambda=-1 \\
& 8 \lambda+7= k -2 \\
& \therefore P =(1,1,-1)
\end{aligned}
\)
Projection of \(2 \hat{i}-2 \hat{k}\) on \(2 \hat{i}+3 \hat{j}+\hat{k}\) is
\(
\begin{aligned}
& =\frac{4-2}{\sqrt{4+9+1}}=\frac{2}{\sqrt{14}} \\
& \therefore l^2=8-\frac{4}{14}=\frac{108}{14} \\
& \Rightarrow 14 l^2=108
\end{aligned}
\)

Hint

Plane \(P \equiv P _1+\lambda P _2=0\)
\(
\begin{aligned}
& (2 x+y-z-3)+\lambda(5 x-3 y)+4 z+9)=0 \\
& (5 \lambda+2) x+(1-3 \lambda) y+(4 \lambda-1) z+9 \lambda-3=0
\end{aligned}
\)
\(\overrightarrow{ n } \cdot \overrightarrow{ b }=0\) where \(\overrightarrow{ b }(2,4,5)\)
\(
\begin{aligned}
& 2(5 \lambda+2)+4(1-3 \lambda)+5(4 \lambda-1)=0 \\
& \lambda=-\frac{1}{6}
\end{aligned}
\)
Plane \(7 x+9 y-10 z-27=0\)

Equation of line \(AB\) is
\(
\frac{ x -8}{-3}=\frac{ y +1}{4}=\frac{z+19}{12}=\lambda
\)
Let \(B=(8-3 \lambda,-1+4 \lambda,-19+12 \lambda)\) lies on plane \(P\)
\(
\begin{aligned}
& \therefore 7(8-3 \lambda)+9(4 \lambda-1)-10(12 \lambda-19)=27 \\
& \lambda=2
\end{aligned}
\)
\(\therefore\) Point \(B=(2,7,5)\)
\(
A B=\sqrt{6^2+8^2+24^2}=26
\)

Hint

Image of point \(\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)\)
\(
\begin{aligned}
& \frac{x-\frac{5}{3}}{1}=\frac{y-\frac{5}{3}}{-2}=\frac{ z -\frac{8}{3}}{1}=\frac{-2\left(1 \times \frac{5}{3}+(-2) \times \frac{8}{3}+1 \times \frac{8}{3}-2\right)}{1^2+2^2+1^2}=\frac{1}{3} \\
& \therefore x=2, y=1, z =3 \\
& P Q^2=13^2=(6-2)^2+(-2-1)^2+(\alpha-3)^2 \\
& \Rightarrow(\alpha-3)^2=144 \Rightarrow \alpha=15(\because \alpha>0)
\end{aligned}
\)

Hint

\(
\begin{aligned}
& A (-6,0,0) \quad B (0,-2,0) C =(0,0,3) \\
& \overrightarrow{ AB }=6 \hat{ i }-2 \hat{ j }, \quad \overrightarrow{ BC }=2 \hat{ j }+3 \hat{ k } \\
& \overrightarrow{ AC }=6 \hat{ i }+3 \hat{ k }
\end{aligned}
\)

\(
\begin{aligned}
& \overrightarrow{ AH } \cdot \overrightarrow{ BC }=0 \\
& \left(\alpha+6, \beta, \frac{6}{7}\right) \cdot(0,2,3)=0 \\
& \beta=\frac{-9}{7} \\
& \overrightarrow{ CH } \cdot \overrightarrow{ AB }=0 \\
& \left(\alpha, \beta, \frac{-15}{7}\right) \cdot(6,-2,0)=0 \\
& 6 \alpha-2 \beta=0 \\
& \alpha=\frac{-3}{7} \\
& 98(\alpha+\beta)^2=(98) \frac{(144)}{49}=288
\end{aligned}
\)

Hint

\(
\begin{aligned}
& L: \frac{x-0}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda} \\
& P: x+2 y+3 z=4
\end{aligned}
\)
Vector parallel to line : \(\langle 1,2, \lambda\rangle=\vec{b}\)
Normal vector to plane \(P:<1,2,3\rangle=\vec{n}\)
Angle between plane and line is \(\theta\)
Then, \(\sin \theta=\frac{<1,2, \lambda>\cdot<1,2,3>}{\sqrt{1^2+2^2+\lambda^2} \cdot \sqrt{1^2+2^2+3^2}}\)
\(
\begin{aligned}
& \Rightarrow \frac{3}{\sqrt{14}}=\frac{1+4+3 \lambda}{\sqrt{\lambda^2+5} \sqrt{14}} \Rightarrow \lambda=\frac{2}{3} \\
& L_1 \equiv \frac{x-0}{3}=\frac{y-1}{6}=\frac{z-3}{2}=\mu
\end{aligned}
\)
Any point on line : \((3 \mu, 6 \mu+1,2 \mu+3)\)
It lies on \(P\)
\(
\begin{aligned}
& \therefore 3 \mu+12 \mu+2+6 \mu+9=4 \\
& \Rightarrow \mu=\frac{-1}{3}
\end{aligned}
\)
Hence, \(\alpha=3 \mu=-1, \beta=6 \mu+1=-1, \gamma=2 \mu+3=\frac{7}{3}\)
Now, \(\alpha+2 \beta+6 \gamma=11\)

Hint

We have,
\(l_1: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in R\) and \(l_2: \vec{r}=(-\hat{\imath}+\hat{ k })+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{ k }), \mu \in R\)
Let direction ratio of line \(l\) be \(a, b\) and \(c\) Equation of line \(l\)
\(
\begin{aligned}
\vec{r} & =(0 \hat{ i }+0 \hat{ j }+0 \hat{ k })+\delta(a \hat{ i }+b \hat{ j }+c \hat{ k }) \\
& =\delta(a \hat{ i }+b \hat{ j }+c \hat{ k })
\end{aligned}
\)
As, line \(l\) is perpendicular to \(l_1\) and \(l_2\),
\(
a \hat{ i }+b \hat{ j }+c \hat{ k }=\left|\begin{array}{ccc}
\hat{ i } & \hat{ j } & \hat{ k } \\
1 & 2 & 3 \\
2 & 2 & 1
\end{array}\right|=-4 \hat{ i }+5 \hat{ j }-2 \hat{ k }
\)
\(\therefore\) Equation of line \(l: \vec{r}=\delta(-4 \hat{ i }+5 \hat{ j }-2 \hat{ k })\)
As, \(P\) is the intersecting point of \(l\) and \(l_1\)
\(
-4 \delta=1+\lambda, 5 \delta=-11+2 \lambda,-2 \delta=-7+3 \lambda
\)
After solving the above three equation, we get
\(
\delta=-1 \text { and } \lambda=3
\)
\(\therefore\) Co-ordinate of point \(P\) is \((4,-5,2)\).
\(Q\) is a point on line \(l_2\)
Let co-ordinate of \(Q\) be \((-1+2 \mu, 2 \mu, 1+\mu)\)
\(
\begin{gathered}
\overrightarrow{P Q}=(-5+2 \mu) \hat{ i }+(2 \mu+5) \hat{ j }+(\mu-1) \hat{ k } \\
\overrightarrow{P Q} \cdot(2 \hat{ i }+2 \hat{ j }+\hat{ k })=0\left[\because \overrightarrow{P Q} \perp l_2\right] \\
2(-5+2 \mu)+2(2 \mu+5)+\mu-1=0 \\
9 \mu-1=0 \Rightarrow \mu=\frac{1}{9}
\end{gathered}
\)
\(
\therefore \alpha=-1+\frac{2}{9}=\frac{-7}{9}, \beta=2 \times \frac{1}{9}=\frac{2}{9}, r=1+\frac{1}{9}=\frac{10}{9}
\)
Hence, \(9(\alpha+\beta+\gamma)=9\left(-\frac{7}{9}+\frac{2}{9}+\frac{10}{9}\right)=5\)

Hint

We have, equation of plane \(P\) is \(x-y+2 z+3=0\)
Perpendicular distance from \(A\) to a plane \(P\) i.e. N
\(
=\frac{|4-3+2(1)+3|}{\sqrt{1^2+(-1)^2+2^2}}
\)
\(B(5, \alpha, \beta)\) lies on the plane \(P\), so
\(
5-\alpha+2 \beta+3=0 \Rightarrow \alpha=2 \beta+8 \dots(i)
\)
Direction ratio of \(A N\) is \(<1,-1,2>\)
Equation of \(A N\) is \(\frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}\)
Co-ordinate of \(N: \frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}=\frac{-(4-3+2+3)}{1+1+4}=-1\)
\(
\Rightarrow x=3, y=4, z=-1
\)
\(\therefore\) Co-ordinate of \(N\) is \((3,4,-1)\)
Also, Area of \((\triangle ABN )=3 \sqrt{2}\)
\(
\begin{aligned}
& \Rightarrow \frac{1}{2} \times AN \times BN =3 \sqrt{2} \\
& \Rightarrow \frac{1}{2} \times \sqrt{6} \times BN =3 \sqrt{2} \Rightarrow BN =2 \sqrt{3}
\end{aligned}
\)
\(
\begin{aligned}
& \text { or } BN ^2=12 \\
& \Rightarrow(5-3)^2+(\alpha-4)^2+(\beta+1)^2=12 \\
& 4+(2 \beta+4)^2+(\beta+1)^2=12(\text { using eqn (i)) } \\
& \Rightarrow 4 \beta^2+16+16 \beta+\beta^2+1+2 \beta=8 \\
& \Rightarrow 5 \beta^2+18 \beta+9=0 \\
& \Rightarrow(5 \beta+3)(\beta+3)=0 \\
& \Rightarrow \beta=-3\left[\because \beta \in z \text {, so rejecting } \beta=\frac{-3}{5}\right] \\
& \text { and } \alpha=2 \beta+8=2(-3)+8=2 \\
& \text { also } \alpha^2+\beta^2+\alpha \beta=9+4-6=7
\end{aligned}
\)

Hint

Equation of plane \(P _2\) passing through \((2,-1,0),(2,0,-1)\) and \((5,1,1)\) is
\(
\begin{aligned}
& \left|\begin{array}{ccc}
x-5 & y-1 & z-1 \\
3 & 2 & 1 \\
3 & 1 & 2
\end{array}\right|=0 \\
& \Rightarrow(x-5)(4-1)-(y-1)(6-3)+(z-1)(3-6)=0 \\
& \Rightarrow 3 x-15-3 y+3-3 z+3=0 \\
& \Rightarrow 3 x-3 y-3 z-9=0 \\
& \Rightarrow x-y-z=3 \ldots \ldots(i)
\end{aligned}
\)
Now, direction ratios of line of intersection of \(P_1\) and \(P _2\) is
\(
\begin{aligned}
& \left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & -1 \\
3 & -1 & -7
\end{array}\right| \\
& =\hat{i}(7-1)-\hat{j}(-7+3)+\hat{k}(-1+3) \\
& =6 \hat{i}+4 \hat{j}+2 \hat{k}
\end{aligned}
\)
At \(z=0, x-y=3\) [from (i)]
\(
3 x-y=11
\)
On solving, we get
\(
x=4 \text { and } y=1
\)
So, equation of line is
\(
\begin{aligned}
& \frac{x-4}{6}=\frac{y-1}{4}=\frac{z-2}{6}=k \\
& \therefore(\alpha, \beta, \gamma)=(6 k+4,4 k+1,2 k) \\
& \Rightarrow(6)(\alpha-7)+4(\beta-4)+2(\gamma+1)=0 \\
& \Rightarrow 6(6 k+4-7)+4(4 k+1-4)+2(2 k+1)=0 \\
& \Rightarrow 36 k-18+16 k-12+4 k+4=0 \\
& \Rightarrow 56 k=26 \Rightarrow k=\frac{1}{2}
\end{aligned}
\)
So, \(\alpha=7, \beta=3\) and \(\gamma=1\)
\(
\therefore \alpha+\beta+\gamma=7+3+1=11
\)

Hint

Since \(\left(\frac{5}{2}, 1, \lambda\right)\) and \((-2,0,1)\) are equidistant
\(
\begin{aligned}
& \text { from plane } 2 x+3 y-6 z+7=0 \\
& \therefore\left|\frac{2\left(\frac{5}{2}\right)+3(1)-6(\lambda)+7}{\sqrt{2^2+3^2+6^2}}\right|=\left|\frac{2(-2)+3(0)-6(1)+7}{\sqrt{2^2+3^2+6^2}}\right| \\
& \Rightarrow|5+3-6 \lambda+7|=|-4-6+7| \\
& \Rightarrow|15-6 \lambda|=|-3| \\
& \Rightarrow 15-6 \lambda= \pm 3 \\
& \Rightarrow 15-6 \lambda=3 \text { or } 15-6 \lambda=-3 \\
& \Rightarrow 6 \lambda=12 \text { or } 6 \lambda=18 \\
& \Rightarrow \lambda=2 \text { or } \lambda=3 \\
& \because \lambda_1>\lambda_2 \\
& \therefore \lambda_1=3 \text { and } \lambda_2=2
\end{aligned}
\)
So, point will be \((1,2,3)\)
Let \(M _0=(1,2,3)\)
\(M_1\) is the point through which line passes i.e, \((5,1,-7)\)
and \(\vec{s}=\hat{i}+2 \hat{j}+2 \hat{k}\)
\(
\therefore \overrightarrow{ M _0 M _1}=4 \hat{i}-\hat{j}-10 \hat{k}
\)
\(
\begin{aligned}
& =\frac{|(4 \hat{i}-\hat{j}-10 \hat{k}) \times(\hat{i}+2 \hat{j}+2 \hat{k})|}{\sqrt{1+4+4}} \\
& =\frac{|18 \hat{i}-18 \hat{j}+9 \hat{k}|}{3}=9
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{\alpha} \\
& \Rightarrow \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\alpha}=\lambda \dots(i)
\end{aligned}
\)
Any point on the line (i)
\(
x=2 \lambda+1, y=3 \lambda+2, z=\alpha \lambda+3
\)
and line \(\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}=\mu \dots(ii)\)
Any point on line (ii)
\(
\Rightarrow x=5 \mu+4, y=2 \mu+1, z=\beta \mu
\)
Since, given lines intersects
\(
\therefore 2 \lambda+1=5 \mu+4 \dots(iii)
\)
\(3 \lambda+2=2 \mu+1 \dots(iv)\)
and \(\alpha \lambda+3=\beta \mu \dots(v)\)
On solving (iii) and (iv), we get
\(
\lambda=-1, \mu=-1
\)
On putting value of \(\lambda\) and \(\mu\) in (v), we get
\(
\begin{gathered}
\alpha(-1)+3=-\beta \\
\Rightarrow \alpha=\beta+3
\end{gathered}
\)
Now,
\(
\begin{aligned}
& 8 \alpha \beta=8(\beta+3)(\beta) \\
& =8\left(\beta^2+3 \beta\right) \\
& =8\left(\beta^2+3 \beta+\frac{9}{4}-\frac{9}{4}\right) \\
& =8\left\{\left(\beta+\frac{3}{2}\right)^2-\frac{9}{4}\right\} \\
& =8\left(\beta+\frac{3}{2}\right)^2-18
\end{aligned}
\)
Here, minimum value \(=-18\)
\(\therefore\) Magnitude of the minimum value of \(8 \alpha \beta\) is 18.

Hint

Let \(Q(x, y, z)\) be the image of \(P(1,2,3)\) in the plane
\(
\begin{aligned}
& 2 x-y+z=9 \\
& \therefore \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{1}=\frac{-2(2 \times 1+(-1)(2)+(1)(3)(-9)}{(2)^2+(-1)^2+(1)^2} \\
& \Rightarrow \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{1}=\frac{-2(-6)}{6}=2 \\
& \Rightarrow x=5, y=0, z=5
\end{aligned}
\)
Now, \(\overrightarrow{P Q}=4 \hat{ i }-2 \hat{ j }+2 \hat{ k }\) and
\(
\overrightarrow{P R}=5 \hat{ i }+8 \hat{ j }+4 \hat{ k }
\)
\(\therefore\) Area of the \(\triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|\)
Now, \(\overrightarrow{P Q} \times \overrightarrow{P R}\)
\(
\begin{aligned}
& =\left|\begin{array}{ccr}
\hat{ i } & \hat{ j } & \hat{ k } \\
4 & -2 & 2 \\
5 & 8 & 4
\end{array}\right| \\
& =\hat{ i }(-8-16)-\hat{ j }(16-10)+\hat{ k }(32+10) \\
& =-24 \hat{ i }-6 \hat{ j }+42 \hat{ k }
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \frac{1}{2}|(-24 \hat{ i }-6 \hat{ j }+42 \hat{ k })|=|-12 \hat{ i }-3 \hat{ j }+21 \hat{ k }| \\
& =\sqrt{(-12)^2+(-3)^2+(21)^2} \\
& =\sqrt{144+9+441}=\sqrt{594} \\
&
\end{aligned}
\)
\(\therefore\) The square of the area of the \(\triangle P Q R=594\)

Hint

Plane: \(8 x+y+2 z=0\)
Given line \(AB : \frac{ x -2}{5}=\frac{ y -4}{10}=\frac{ z +3}{-4}=\lambda\)
Any point on line \((5 \lambda+2,10 \lambda+4,-4 \lambda-3)\)
Point of intersection of line and plane
\(
\begin{aligned}
& 8(5 \lambda+2)+10 \lambda+4-8 \lambda-6=0 \\
& \lambda=-\frac{1}{3} \\
& C \left(\frac{1}{3}, \frac{2}{3},-\frac{5}{3}\right)
\end{aligned}
\)
\(L : \frac{ x -1}{-1}=\frac{ y +4}{2}=\frac{ z +2}{3}=\mu\)
\(
\begin{aligned}
& \overrightarrow{ CD }=\left(-\mu+\frac{2}{3}\right) \hat{ i }+\left(2 \mu-\frac{14}{3}\right) \hat{ j }+\left(3 \mu-\frac{1}{3}\right) \hat{ k } \\
& \left(-\mu+\frac{2}{3}\right)(-1)+\left(2 \mu-\frac{14}{3}\right) 2+\left(3 \mu-\frac{1}{3}\right) 3=0 \\
& \Rightarrow \mu=\frac{11}{14} \\
& \overrightarrow{ CD }=\frac{-5}{42}, \frac{-130}{42}, \frac{85}{42}
\end{aligned}
\)
Direction ratios \(\rightarrow(-1,-26,17)\)
\(
|a+b+c|=10
\)

Hint

Plane :
\(
a(x-3)+b(y+2)+c(z-5)=0
\)
Dr’s of plane : \(3 \hat{i}-\hat{j}-2 \hat{k}\)
\(
\begin{aligned}
& <3,-1,-2> \\
& P: 3(x-3)-1(y+2)-2(z-5)=0 \\
& 3 x-9-y-2-2 z+10=0 \\
& 3 x-y-2 z=1
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \alpha=3, \beta=-1, \gamma=-2 \\
& \Rightarrow \alpha \beta \gamma=6
\end{aligned}
\)

Hint

\(
\begin{aligned}
& L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}=r \text { (say) } \\
& \text { Let } P \equiv\left(2 r_1+1,-r_1, r_1+3\right) \\
& P \text { lies on } 2 x+y+3 z=16 \\
& \therefore 2\left(2 r_1+1\right)+\left(-r_1-1\right)+3\left(r_1+3\right)=16 \\
& r_1=1 \\
& P \equiv(3,-2,4) \\
& R \equiv(1,-1,-3) \\
& \text { Let } Q \equiv\left(2 r_2+1,-r_2-1, r_2+3\right) \\
& D R \text { of } Q R \equiv\left(2 r_2-r_2 r_2+6\right) \\
& \text { DRs of } L \equiv(2,-1,1) \\
& Q R \perp L \Rightarrow 4 r_2+r_2+r_2+6=0 \\
& r_2=-1 \\
& Q \equiv(-1,0,2)
\end{aligned}
\)
\(
\begin{aligned}
& \overrightarrow{Q P} \times \overrightarrow{R P}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
4 & -2 & 2 \\
2 & -1 & 7
\end{array}\right|=-12 \hat{i}-24 \hat{j}+0 \hat{k} \\
& \alpha=[P Q R]=\frac{1}{2}|\overrightarrow{Q P} \times \overrightarrow{R P}|=\frac{1}{2} \times 12 \sqrt{5} \\
& =6 \sqrt{5} \\
& \alpha^2=180
\end{aligned}
\)

Hint

\(
\begin{aligned}
& P_1: \vec{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})=9 \\
& P_2: \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=15 \\
& \text { then } \cos \theta=\frac{3}{\sqrt{6} \cdot \sqrt{6}}=\frac{1}{2} \\
& \therefore \theta=\frac{\pi}{3}, \text { Now } \alpha=\frac{\pi}{2}-\theta \\
& \therefore \tan ^2 \theta \cdot \cot ^2 \alpha=\tan ^4 \theta \\
& =(\sqrt{3})^4=9
\end{aligned}
\)

Hint

\(
L: \frac{x-2}{1}=\frac{y-3}{-1}=\frac{z-1}{-1}=\lambda
\)
Any point on \(L\) can be taken as
\(
B(\lambda+2,-\lambda+3,-\lambda+1)
\)
Let \(A(5,3,8)\)
\(
\begin{aligned}
& \text { So, } A B \cdot(\hat{i}-\hat{j}-\widehat{k})=0 \\
& {[(\lambda-3) \hat{i}-\lambda \hat{j}-(\lambda+7) \hat{k}] \cdot[\hat{i}-\hat{j}-\widehat{k}]=0} \\
& \lambda-3+\lambda+\lambda+7=0 \\
& \therefore \lambda=\frac{-4}{3} \\
& \overrightarrow{A B}=\frac{13}{3} \hat{i}+\frac{4}{3} \hat{i}-\frac{17}{3} \widehat{k} \\
& |\overrightarrow{A B}|=\sqrt{\frac{169}{9}+\frac{16}{9}+\frac{289}{9}} \\
& =\frac{\sqrt{474}}{3}=\alpha \\
& 3 \alpha^2=\frac{474}{9} \times 3=158
\end{aligned}
\)

Hint

Line of intersection of the planes \(x-3 y+2 z-1=0\) and \(4 x-y+z=0\) is normal \((\vec{n})\) to the required plane.
\(
\vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
1 & -3 & 2 \\
4 & -1 & 1
\end{array}\right|=-\hat{i}+7 \hat{j}+11 \widehat{k}
\)
Equation of plane is
\(
-x+7 y+11 z=\lambda
\)
It passes through \((1,1,2)\)
\(
\therefore \lambda=28
\)
So, the plane is
\(
\begin{aligned}
& -x+7 y+11 z=28 \\
& \Rightarrow \frac{-1}{28} x+\frac{7}{28} y+\frac{11}{28} z=1 \\
& A=\frac{-1}{28}, B=\frac{7}{28}, C=\frac{11}{28} \\
& 140(C-B+A)=15
\end{aligned}
\)

Hint

\(
\begin{aligned}
& P_1: \vec{r} \cdot(3 \hat{i}-5 \hat{j}+\widehat{k})=7 \\
& P_2: \vec{r} \cdot(\lambda \hat{i}+\hat{j}-3 \widehat{k})=9
\end{aligned}
\)
Let angle between \(P_1\) and \(P_2\) is \(\theta\)
Then \(\cos \theta=\frac{3 \lambda-5-3}{\sqrt{35} \sqrt{\lambda^2+10}}\)
\(
\begin{aligned}
& \text { But } \sin \theta=\frac{2 \sqrt{6}}{5} \\
& \therefore \frac{(3 \lambda-8)^2}{35\left(\lambda^2+10\right)}=1-\frac{24}{25} \\
& \Rightarrow 5\left(9 \lambda^2+64-48 \lambda\right)=7 \lambda^2+70 \\
& \Rightarrow 38 \lambda^2-240 \lambda+250=0 \\
& \Rightarrow 19 \lambda^2-120 \lambda+125=0 \\
& \Rightarrow(19 \lambda-25)(\lambda-5)=0 \\
& \therefore \lambda_1=\frac{25}{19}, \lambda_2=5
\end{aligned}
\)
So, point \((50,50,2)\)
\(
\therefore d=\frac{|150-250+2-7|}{\sqrt{35}}=315
\)

Hint

The line \(\frac{x+10}{1}=\frac{y-8}{-2}=\frac{z}{1}\) have a point \((-10,8,0)\) with d. r. \((1,-2,1)\)
\(\because\) the plane \(a x+b y+3 z=2(a+b)\)
\(
\Rightarrow b =2 a
\)
\(\&\) dot product of d.r.’s is zero
\(
\therefore a-2 b+3=0 \therefore a=1 \& b=2
\)
Distance from \((1,27,7)\) is
\(
\begin{aligned}
& c=\frac{1+54+21-6}{\sqrt{14}}=\frac{70}{\sqrt{14}}=5 \sqrt{14} \\
& \therefore a ^2+ b ^2+ c ^2=1+4+350 \\
& =355
\end{aligned}
\)

Hint

\(
\left|\frac{1}{2} \cdot 2 \sqrt{21} \cdot\right| \begin{array}{ccc}
i & j & k \\
\alpha & 1 & \alpha+4 \\
5 & 2 & 3
\end{array}\left|\frac{1}{\sqrt{25+4+9}}\right|=21
\)
\(
\begin{aligned}
& \sqrt{(2 \alpha+5)^2+(2 \alpha+20)^2+(2 \alpha-5)^2}=\sqrt{21} \sqrt{38} \\
& \Rightarrow 12 \alpha^2+80 \alpha+450=798 \\
& \Rightarrow 12 \alpha^2+80 \alpha-348=0 \\
& \Rightarrow \alpha=3 \Rightarrow \alpha^2=9 \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Points }(1,2,3) \text { and }(2,3,4) \\
& L_1: \frac{(x-1)}{1}=\frac{(y-2)}{1}=\frac{(2-3)}{1} \\
& L_2: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0} \\
& \vec{b}_1=\hat{i}+\hat{j}+\hat{k} \\
& \vec{b}_2=2 \hat{i}-\hat{j}+0 \hat{k} \\
& \overrightarrow{a_1}-\overrightarrow{a_2}=0 \hat{i}-3 \hat{j}-\hat{k} \\
& d=\left|\frac{\left(\bar{a}_1-\bar{a}_2\right) \cdot\left(n_1 \times n_2\right)}{\left|n_1 \times n_2\right|}\right| \\
& =\left|\frac{6-3}{\sqrt{9+1+4}}\right|=\frac{3}{\sqrt{14}}=\alpha \\
& 28 \alpha^2=\frac{28 \times 9}{14}=18
\end{aligned}
\)

Hint

Let the equation of the plane is
\(
(x-2 y-z-5)+\lambda(x+y+3 z-5)=0 \dots(i)
\)
\(\because \quad\) it’s parallel to the line
\(
x+y+2 z-7=0=2 x+3 y+z-2
\)
So, vector along the line \(\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{array}\right|\)
\(
=-5 \hat{i}+3 \hat{j}+\hat{k}
\)
\(\because\) Plane is parallel to line
\(
\begin{aligned}
& \therefore-5(1+\lambda)+3(-2+\lambda)+1(-1+3 \lambda)=0 \\
& \lambda=12
\end{aligned}
\)
So, by (i)
\(
\begin{aligned}
& 13 x+10 y+35 z=65 \\
& \therefore a=13, b=10, c=35 \text { and } d=\frac{26+20-35+16}{\sqrt{9}}=9
\end{aligned}
\)

Hint

Shortest distance between the lines
\(
\begin{aligned}
& \frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4} \\
& \frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5} \text { is } 6
\end{aligned}
\)
Vector along line of shortest distance
\(=\left|\begin{array}{lll} i & j & k \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|, \Rightarrow-\hat{ i }+2 \hat{ j }- k\) (its magnitude is \(\sqrt{6}\) )
Now \(\frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\sqrt{6}+\lambda & \sqrt{6} & -3 \sqrt{6} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|= \pm 6\)
\(
\Rightarrow \lambda=-2 \sqrt{6}, 10 \sqrt{6}
\)
So, square of sum of these values
\(
=(10 \sqrt{6}-2 \sqrt{6})^2=(8 \sqrt{6})^2=384
\)

Hint

Shortest distance between the lines
\(
\begin{aligned}
& =\frac{\left|\begin{array}{ccc}
4 & 2 & -14 \\
3 & 2 & 2 \\
3 & -2 & 0
\end{array}\right|}{\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 2 & 2 \\
3 & -2 & 0
\end{array}\right|} \\
& =\frac{16+12+168}{|4 \hat{i}+6 \hat{j}-12 \hat{k}|}=\frac{196}{14}=14
\end{aligned}
\)

Hint

Given \(a \cdot 3+(-4 a)(-1)+(-7) 2 b=0 \dots(i)\)
and \(a b-4 a^2+14=0 \dots(ii)\)
\(\Rightarrow a^2=4\) and \(b^2=1\)
\(\therefore L \equiv \frac{x+1}{5}=\frac{y-2}{3}=\frac{z}{1}=\lambda\) (say)
\(\Rightarrow\) General point on line is \((5 \lambda-1,3 \lambda+2, \lambda)\) for finding point of intersection with \(x-y+z=0\) we get \((5 \lambda-1)-(3 \lambda+2)+(\lambda)=0\)
\(\Rightarrow 3 \lambda-3=0 \Rightarrow \lambda=1\)
\(\therefore\) Point at intersection \((4,5,1)\)
\(
\therefore \alpha+\beta+\gamma=4+5+1=10
\)

Hint

\(
\begin{aligned}
& \text { d.r’s of } R S=<\alpha,-1, \beta> \\
& \text { d.r’s of } P Q=<\frac{90}{17}, \frac{60}{17}, \frac{94}{17}>=<45,30,47>
\end{aligned}
\)
as \(P Q\) and \(R S\) are diagonals of rhombus
\(
\begin{aligned}
& \alpha(45)+30(-1)+47(\beta)=0 \\
& \Rightarrow 45 \alpha+47 \beta=30 \\
& \text { i.e., } \alpha=\frac{30-47 \beta}{45}
\end{aligned}
\)
for minimum integral value \(\alpha=-15\) and \(\beta=15\)
\(
\Rightarrow \alpha^2+\beta^2=450
\)

Hint

Equation of plane containing the line \(4 a x-y+5 z-7 a=0=2 x-5 y-z-3\) can be written as
\(
\begin{aligned}
& 4 a x-y+5 z-7 a+\lambda(2 x-5 y-z-3)=0 \\
& (4 a+2 \lambda) x-(1+5 \lambda) y+(5-\lambda) z-(7 z+3 \lambda)=0
\end{aligned}
\)
Which is coplanar with the line
\(
\begin{aligned}
& \frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1} \\
& 4(4 a+2 \lambda)+(1+5 \lambda)-(7 a+3 \lambda)=0 \\
& 9 a+10 \lambda+1=0 \ldots . .(1) \\
& (4 a+2 \lambda) 1+(1+5 \lambda) 2+5-\lambda=0 \\
& 4 a+11 \lambda+7=0 \ldots \ldots .(2) \\
& a=1, \lambda=-1
\end{aligned}
\)
Equation of plane is \(x+2 y+3 z-2=0\)
Intersection with the line
\(
\begin{aligned}
& \frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4} \\
& (7 t+3)+2(-t+2)+3(-4 t+3)-2=0 \\
& -7 t+14=0 \\
& t=2
\end{aligned}
\)
So, the required point is \((17,0,-5)\)
\(
\alpha+\beta+\gamma=12
\)

Hint

\(
\begin{aligned}
& \text { Normal to plane }=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
1 & a & -1 \\
-1 & 1 & -a
\end{array}\right| \\
& =\hat{i}\left(1-a^2\right)-\hat{j}(-a-1)+\widehat{k}(1+a) \\
& =(1-a) \hat{i}+\hat{j}+\widehat{k} \\
& \therefore \text { Plane }(1-a)(x-1)+(y-1)+z=0
\end{aligned}
\)
Distance from \((2,1,4)\) is \(\sqrt{3}\) i.e.
\(
\begin{aligned}
& \Rightarrow\left|\frac{(1-a)+0+4}{\sqrt{(1-a)^2+1+1}}\right|=\sqrt{3} \\
& \Rightarrow 25+a^2-10 a=3 a^2-6 a+9 \\
& \Rightarrow 2 a^2+4 a-16=0 \\
& \Rightarrow a^2+2 a-8=0 \\
& a=2 \text { or }-4 \\
& \therefore a_{\max }=2
\end{aligned}
\)

Hint

\(L: l x-y+3(1-l) z=1, x+2 y-z=2\) and plane containing the line
\(
p: 3 x-8 y+7 z=4
\)
Let \(\vec{n}\) be the vector parallel to \(L\).
\(
\begin{aligned}
& \text { then } \vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
l & -1 & 3(1-l) \\
1 & 2 & -1
\end{array}\right| \\
& =(6 l-5) \hat{i}+(3-2 l) \hat{j}+(2 l+1) \hat{k}
\end{aligned}
\)
\(\because R\) containing \(L\)
\(
\begin{aligned}
& 3(6 l-5)-8(3-2 l)+7(2 l+1)=0 \\
& 18 l+16 l+14 l-15-24+7=0 \\
& \therefore l=\frac{32}{48}=\frac{2}{3}
\end{aligned}
\)
Let \(\theta\) be the acute angle between \(L\) and \(y\)-axis
\(
\begin{aligned}
& \therefore \cos \theta=\frac{\frac{5}{3}}{\sqrt{1+\frac{25}{9}+\frac{49}{9}}}=\frac{5}{\sqrt{83}} \\
& \therefore 415 \cos ^2 \theta=125
\end{aligned}
\)

Hint

\(
L: \frac{x+1}{2}=\frac{y+2}{3}=\frac{2-1}{2}
\)
\(
\begin{aligned}
& \text { Let } T(2 t-1,3 t-2,2 t+1) \\
& \because P T \perp^r Q R \\
& \therefore 2(2 t-5)+3(3 t-4)+2(2 t-6)=0 \\
& 17 t=34 \\
& \therefore t=2
\end{aligned}
\)
So \(T(3,4,5)\)
\(
\begin{aligned}
& \therefore P T=\sqrt{1+4+4}=3 \\
& \therefore Q T=\sqrt{26-9}=\sqrt{17}
\end{aligned}
\)
\(\therefore\) Area of \(\triangle P Q R=\frac{1}{2} \times 2 \sqrt{17} \times 3=3 \sqrt{17}\)
\(\therefore\) Square of \(\operatorname{ar}(\triangle P Q R)=153\).

Hint

DR’s of line of shortest distance
\(
\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 1 & 1 \\
2 & 2 & 1
\end{array}\right|=-\hat{i}+2 \hat{j}-2 \hat{k}
\)
angle between line and plane is \(\cos ^{-1} \sqrt{\frac{2}{27}}=\alpha\)
\(
\cos \alpha=\sqrt{\frac{2}{27}}, \sin \alpha=\frac{5}{3 \sqrt{3}}
\)
DR’s normal to plane \((1,-1,-1)\)
\(
\begin{aligned}
& \sin \alpha=\left|\frac{-a-2+2}{\sqrt{4+4+1} \sqrt{a^2+1+1}}\right|=\frac{5}{3 \sqrt{3}} \\
& \sqrt{3}|a|=5 \sqrt{a^2+2} \\
& 3 a^2=25 a^2+50
\end{aligned}
\)
No value of (a)

Hint

Centroid of \(\triangle ABC =G\left(\frac{2 \alpha}{3}, \frac{2 \alpha}{3}, \frac{2 \alpha}{3}\right)\)
Given equation of line is \(\frac{x}{1}=\frac{z-3}{-1}=\frac{y}{0}=\lambda\)
\(
x=\lambda, y=0, z=-\lambda+3
\)
\(\therefore D(\lambda, 0-\lambda+3)\) be any point on given line
\(
\begin{aligned}
& \therefore G D=\sqrt{\left(\lambda-\frac{2 \alpha}{3}\right)^2+\left(\frac{2 \alpha}{3}\right)^2+\left(-\lambda+3-\frac{2 \alpha}{3}\right)^2} \\
& GD _1=\left(\lambda-\frac{2 \alpha}{3}\right)^2+\left(\frac{2 \alpha}{3}\right)^2+\left(-\lambda+3-\frac{2 \alpha}{3}\right)^2 \\
& \frac{d G D_1}{d \lambda}=2\left(\lambda-\frac{2 \alpha}{3}\right)-2\left(-\lambda+3-\frac{2 \alpha}{3}\right) \\
& =4 \lambda-6=0 \\
& \Rightarrow \lambda=\frac{3}{2}
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \text { Minimum GD }=\sqrt{\left(\frac{3}{2}-\frac{2 \alpha}{3}\right)^2+\left(\frac{2 \alpha}{3}\right)^2+\left(-\frac{3}{2}+3-\frac{2 \alpha}{3}\right)^2} \\
& \sqrt{\frac{57}{2}}=\sqrt{\left(\frac{9-4 \alpha}{6}\right)^2+\frac{4 \alpha^2}{9}+\left(\frac{9-4 \alpha}{6}\right)^2} \\
& \Rightarrow \frac{57}{2}=\frac{24 \alpha^2-72 \alpha+81}{18} \\
& \Rightarrow \alpha^2-3 \alpha-18=0 \ldots(\because \alpha>0) \\
& \Rightarrow \alpha=-3,6 \\
& \therefore \alpha=6
\end{aligned}
\)

Hint

Foot of perpendicular from \(P\)
\(
\begin{aligned}
& \frac{x-1}{-1}=\frac{y-2}{1}=\frac{z+1}{1}=\frac{-(-1+2-1-1)}{3} \\
& \Rightarrow p^{\prime} \equiv\left(\frac{2}{3}, \frac{7}{3}, \frac{-2}{3}\right)
\end{aligned}
\)
and foot of perpendicular from \(Q\)
\(
\begin{aligned}
& \frac{x-2}{-1}=\frac{y+1}{1}=\frac{z-3}{1}=\frac{-(-2-1+3-1)}{3} \\
& \Rightarrow Q^{\prime} \equiv\left(\frac{5}{3}, \frac{-2}{3}, \frac{10}{3}\right) \\
& P^{\prime} Q^{\prime}=\sqrt{(1)^2+(3)^2+(4)^2}=d=\sqrt{26} \\
& \Rightarrow d^2=26
\end{aligned}
\)

Hint

Direction ratio of normal to \(\left.P_1 \equiv<2,1,-3\right\rangle\) and that of \(P_2 \equiv\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 0 & 1 & -5 \\ -1 & -2 & 5\end{array}\right|=-5 \hat{i}-\hat{j}(-5)+\widehat{k}(1)\)
i.e. \(\langle-5,5,1\rangle\)
d.r’s of line of intersection are along vector
\(
\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \widehat{k} \\
2 & 1 & -3 \\
-5 & 5 & 1
\end{array}\right|=\hat{i}(16)-\hat{j}(-13)+\widehat{k}(15)
\)
i.e. \(\langle 16,13,15\rangle\)
\(
\therefore \alpha+\beta=13+15=28
\)

Hint

The point dividing \(P Q\) in the ratio \(1: 3\) will be mid-point of \(P \&\) foot of perpendicular from \(P\) on the line.
\(\therefore\) Let a point on line be \(\lambda\)
\(
\begin{aligned}
& \Rightarrow \frac{x-6}{3}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda \\
& \Rightarrow P^{\prime}(3 \lambda+6,2 \lambda+1,3 \lambda+2)
\end{aligned}
\)
as \(P ^{\prime}\) is foot of perpendicular
\(
\begin{aligned}
& (3 \lambda+5) 3+(2 \lambda-1) 2+(3 \lambda-1) 3=0 \\
& \Rightarrow 22 \lambda+15-2-3=0 \\
& \Rightarrow \lambda=\frac{-5}{11} \\
& \therefore P^{\prime}\left(\frac{51}{11}, \frac{1}{11}, \frac{7}{11}\right)
\end{aligned}
\)
Mid-point of \(P P^{\prime} \equiv\left(\frac{\frac{51}{11}+1}{2}, \frac{\frac{1}{11}+2}{2}, \frac{\frac{7}{11}+3}{2}\right)\)
\(
\begin{aligned}
& \equiv\left(\frac{62}{22}, \frac{23}{22}, \frac{40}{22}\right) \equiv(\alpha, \beta, \gamma) \\
& \Rightarrow 22(\alpha, \beta, \gamma)=62+23+40=125
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-2(3 a-4 b+12 c+19)}{3^2+(-4)^2+12^2} \\
& \frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-6 a+8 b-24 c-38}{169} \\
& (x, y, z) \equiv(a-6, \beta, \gamma) \\
& \frac{(a-6)-a}{3}=\frac{\beta-b}{-4}=\frac{\gamma-c}{12}=\frac{-6 a+8 b-24 c-38}{169} \\
& \frac{\beta-b}{-4}=-2 \Rightarrow \beta=8+b \\
& \frac{\gamma-c}{12}=-2 \Rightarrow \gamma=-24+c \\
& \frac{-6 a+8 b-24 c-38}{169}=-2
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow 3 a-4 b+12 c=150 \ldots . .(1)\\
&\begin{aligned}
& a+b+c=5 \\
& 3 a+3 b+3 c=15 \ldots \ldots .(2)
\end{aligned}
\end{aligned}
\)
Applying (1) – (2)
\(
\begin{aligned}
& -7 b+9 c=135 \\
& 7 b-9 c=-135 \\
& 7 \beta-9 \gamma=7(8+b)-9(-24+c) \\
& =56+216+7 b-9 c \\
& =56+216-135=137
\end{aligned}
\)

Hint

Equation of \(l_1\)
\(
\frac{x-\frac{1}{8}}{\frac{1}{8}}=\frac{y}{-\frac{1}{4 \sqrt{2}}}=\frac{z}{0}
\)
\(
\text { or } \frac{x-\frac{1}{8}}{1}=\frac{y}{-\sqrt{2}}=\frac{z}{0} \ldots . . \text { (i) }
\)
Equation of \(L _2\)
\(
\begin{aligned}
& \frac{x+\frac{1}{8}}{-6 \sqrt{3}}=\frac{y}{0}=\frac{z}{8} \dots(ii) \\
& d=\left|\frac{(\vec{c}-\vec{a}) \cdot \vec{b} \times \vec{d})}{|\vec{b} \times \vec{d}|}\right| \\
& =\frac{\left(\frac{1}{4} \hat{i}\right) \cdot(4 \sqrt{2} \hat{i}+4 \hat{j}+3 \sqrt{6 \hat{k}})}{\sqrt{(4 \sqrt{2})^2+4^2+(3 \sqrt{6})^2}} \\
& =\frac{\sqrt{2}}{\sqrt{32+16+54}}=\frac{1}{\sqrt{51}} \\
& d^{-2}=51 \\
&
\end{aligned}
\)

Hint

Both the lines lie in the same plane

\(\therefore\) equation of the plane
\(
\begin{aligned}
& \left|\begin{array}{lll}
x & y & z \\
1 & 2 & 3 \\
1 & 1 & 5
\end{array}\right|=0 \\
& \Rightarrow 7 x -2 y – z =0 \\
& \therefore a + b + d =5
\end{aligned}
\)

Hint

Let \(A(3 \lambda+7,-\lambda+1, \lambda-2)\) and \(B(2 \mu, 3 \mu+7, \mu)\)
So, DR’s of \(A B \propto 3 \lambda-2 \mu+7,-(\lambda+3 \mu+6), \lambda-\mu-2\)
Clearly \(\frac{3 \lambda-2 \mu+7}{1}=\frac{\lambda+3 \mu+6}{4}=\frac{\lambda-\mu-2}{2}\)
\(
\Rightarrow 5 \lambda-3 \mu=-16 \dots(i)
\)
And \(\lambda-5 \mu=10 \dots(ii)\)
From (i) and (ii) we get \(\lambda=-5, \mu=-3\)
So, \(A\) is \((-8,6,-7)\) and \(B\) is \((-6,-2,-3)\)
\(
A B=\sqrt{4+64+16} \Rightarrow(A B)^2=84
\)

Hint

\(
\begin{aligned}
& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -a & 0 \\
1 & -1 & 1
\end{array}\right|=-a \hat{i}-\hat{j}+(a-1) \hat{k} \\
& \vec{a}_1-\vec{a}_2=-\hat{i}+\hat{j}+\hat{k} \\
& \text { Shortest distance }=\left|\frac{\left(\vec{a}_1-\vec{a}_2\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right| \\
& \Rightarrow \quad \sqrt{\frac{2}{3}}=\frac{2(a-1)}{\sqrt{a^2+1+(a-1)^2}} \\
& \Rightarrow 6\left(a^2-2 a+1\right)=2 a^2-2 a+2 \\
& \Rightarrow \quad(a-2)(2 a-1)=0 \Rightarrow a=2 \text { because } a \in z .
\end{aligned}
\)

Hint

Given equation of line
\(
\frac{x-2}{\alpha}=\frac{y-2}{-5}=\frac{z+2}{2} \ldots \ldots \text { (i) }
\)
and plane \(x+3 y-2 z+\beta=0 \dots(ii)\)
Line (i) passes through \((2,2,-2)\)
which lies on plane (ii).
\(
\therefore 2+6+4+\beta=0 \Rightarrow \beta=-12
\)
Also, given line is perpendicular to normal of the plane
\(
\begin{aligned}
& \alpha(1)-5(3)+2(-2)=0 \Rightarrow \alpha=19 \\
& \therefore \alpha+\beta=19+(-12)=19-12=7
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}=\lambda \\
& x=2 \lambda+1, y=3 \lambda+2, z=6 \lambda-1
\end{aligned}
\)
for point of intersection of line & plane
\(
\begin{aligned}
& 2(2 \lambda+1)-(3 \lambda+2)+(6 \lambda-1)=6 \\
& 7 \lambda=7 \Rightarrow \lambda=1
\end{aligned}
\)
point : \((3,5,5)\)
\(
\begin{aligned}
& (\text { distance })^2=(3+1)^2+(5+1)^2+(5-2)^2 \\
& =16+36+9=61
\end{aligned}
\)

Hint

Since \(R(3,5, \gamma)\) lies on the plane \(2 x-y+z+3=0\).
Therefore, \(6-5+\gamma+3=0\)
\(
\Rightarrow \gamma=-4
\)
Now,
dr’s of line QS are \(2,-1,1\)
equation of line QS is
\(
\begin{aligned}
& \frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\lambda \text { (say) } \\
& \Rightarrow F(2 \lambda+1,-\lambda+3, \lambda+4)
\end{aligned}
\)
F lies in the plane
\(
\begin{aligned}
& \Rightarrow 2(2 \lambda+1)-(-\lambda+3)+(\lambda+4)+3=0 \\
& \Rightarrow 4 \lambda+2+\lambda-3+\lambda+7=0 \\
& \Rightarrow 6 \lambda+6=0 \Rightarrow \lambda=-1 \\
& \Rightarrow F(-1,4,3)
\end{aligned}
\)
Since, \(F\) is mid-point of QS.
Therefore, coordinated of \(S\) are \((-3,5,2)\).
So, \(S R=\sqrt{36+0+36}=\sqrt{72}\)
\(
SR ^2=72 \text {. }
\)