JEE Main 2020 — Three Dimensional Geometry Question with Solution

Equation of plane of passing through intersection of planes $P_1\Rightarrow x+4y-z+7=0$        ......$\left(1\right)$ and $P_2\Rightarrow 3x+y+5z-8=0$      .....$\left(2\right)$

is $P_1+\lambda P_2=0$

$\left(x+4y-z+7\right)+\lambda \left(3x+y+5z-8\right)=0$

$\left(3\lambda +1\right)x+\left(\lambda +4\right)y+\left(5\lambda -1\right)z+\left(7-8\lambda \right)=0$     .......$\left(3\right)$

$ax+by+6z-15=0$               ........$\left(4\right)$

On comparing $\left(3\right)$ and $\left(4\right)$,

$\frac{a}{3\lambda +1}=\frac{b}{\lambda +4}=\frac{6}{5\lambda -1}=\frac{-15}{7-8\lambda }$

$\therefore$ $42-48\lambda =-75\lambda +15$

 $27\lambda =-27$

 $\lambda =-1$

$\therefore$ $\frac{a}{-2}=\frac{b}{3}=-1$

$\Rightarrow$ $a=2, b=-3$

 So, equation of plane is  $2x-3y+6z-15=0$         ......$\left(5\right)$

Distance of plane $\left(5\right)$ from point $\left(3, 2, -1\right)$ is $p=\left|\frac{6-6-6-15}{\sqrt{4+9+36}}\right|=\left|-\frac{21}{\sqrt{49}}\right|=3$.