JEE Main 2023 — Three Dimensional Geometry Question with Solution

Given,

Equation of planes $P_1: x+\left(\lambda +4\right)y+z=1$ and $P_2: 2x+y+z=2$ 

Now we know that equation of plane passing through the planes $P_1 \& P_2$ is given by $P_1+k{P}_2=0$

Now given it is passing through the points $\left(0,1,0\right)$ and $\left(1,0,1\right)$,

So, $x+\left(\lambda +4\right)y+z-1+k\left(12x+y+z-2\right)=0$ when passing through $\left(0,1,0\right)$ we get,

$0+\left(\lambda +4\right)1+0-1+k\left(12\times 0+1+0-2\right)=0$

$\Rightarrow \lambda +4-1-k=0$\

$\Rightarrow \lambda +3-k=0 ........\left(1\right)$

And when passing through $\left(1,0,1\right)$ we get, $k=-1$

So, from equation $\left(1\right)$ we get, $\lambda =-4$

So, point $\left(2\lambda ,\lambda ,-\lambda \right)$ will be $\left(-8,-4,4\right)$

And distance of point$\left(-8,-4,4\right)$  from the plane $P_2$ will be,

$D=\left|\frac{-2\times 8-1\times 4+1\times 4-2}{\sqrt{2^2+1^2+1^2}}\right|=\frac{18}{\sqrt{6}}=3\sqrt{6}$