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  4. A plane E is perpendicular to the two planes 2 x-2 y+z=0 and x-y+2 z=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 √2, then ( PQ )2 is equal to

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Q. A plane $E$ is perpendicular to the two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, and passes through the point $P (1,-1,1)$. If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3 \sqrt{2}$, then $( PQ )^2$ is equal to

JEE MainJEE Main 2022Three Dimensional Geometry

Solution:

Let equation of plane be $a ( x -1)+ b ( y +1)+ c ( z -1)=0$.....(1)
It is perpendicular to the given two planes
$ 2 a-2 b+c=0$
$ a-b+2 c=0$
$\Rightarrow \frac{a}{3}=\frac{b}{3}=\frac{c}{0}$
Equation of plane be $x + y =0$
Now $\frac{| a + a |}{\sqrt{2}}=3 \sqrt{2} \Rightarrow|2 a |=6 \Rightarrow a =\pm 3$
$P (3,3,2)$ or $P (-3,-3,2), Q (1,-1,1)$
$PQ ^2=(3-1)^2+(3+1)^2+(2-1)^2=21$