JEE Main 2022 (Online) 25th July Evening ShiftQuestion 1
A plane $E$ is perpendicular to the two planes $2x-2y+z=0$ and $x-y+2z=4$, and passes through the point $P(1,-1,1)$. If the distance of the plane $E$ from the point $Q(a,a,2)$ is $3\sqrt2$, then $(PQ)^2$ is equal to
3D Geometry
Solution
Normal of $E$: $(2,-2,1)\times(1,-1,2)=(-3,-3,0)\parallel(1,1,0)$. Through $P(1,-1,1)$: $x+y=0$. Distance of $Q(a,a,2)$: $\frac{|2a|}{\sqrt2}=3\sqrt2\Rightarrow|a|=3$. $PQ^2=(a-1)^2+(a+1)^2+1=2a^2+3=21$. Option (C).