JEE Main 2022 — Three Dimensional Geometry Question with Solution

Given the plane passing through the line $L_{1 }:l x-y+3\left(1-l\right)z=1$, $L_2: x+2y-z=2$ and perpendicular to the plane $3x+2y+z=6$ is $3x-8y+7z=4$,

So direction ratio of line $L_1$ will be ${\overrightarrow{n}}_1=l\hat{i}-\hat{j}+3\left(1-l\right)\hat{k}$

And $L_2$ will be ${\overrightarrow{n}}_2=\hat{i}+2\hat{j}-\hat{k}$

Now direction ratio of line $=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ l& -1& 3\left(1-l\right)\\ 1& 2& -1\end{array}\right|$

$=\left(6l-5\right)\hat{i}+\left(3-2l\right)\hat{j}+\left(2l+1\right)\hat{k}$

Also plane $3x-8y+7z=4$ will contain the line $\left(6l-5\right)\hat{i}+\left(3-2l\right)\hat{j}+\left(2l+1\right)\hat{k}$

So, normal of $3x-8y+7z=4$ will be perpendicular to the line

$\Rightarrow 3\left(6l-5\right)+\left(3-2l\right)\left(-8\right)+7\left(2l+1\right)=0$

$\Rightarrow l=\frac{2}{3}$

So, direction ratio of line $\left(-1,\frac{5}{3},\frac{7}{3}\right)$

Now finding angle with $y$ axis

$\cos \theta =\frac{{\displaystyle \frac{5}{3}}}{\sqrt{1+\frac{25}{9}+\frac{49}{9}}}$

$\cos \theta =\frac{5}{\sqrt{83}}$

So, $415{\cos }^2\theta =\frac{25}{83}\times 415=125$