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 $(P Q)^{2}$ is equal to :

<p>First plane, ${P_1} = 2x - 2y + z = 0$, normal vector $\equiv {\overline n _1} = (2, - 2,1)$</p> <p>Second plane, ${P_2} \equiv x - y + 2z = 4$, normal vector $\equiv {\overline n _2} = (1, - 1,2)$</p> <p>Plane perpendicular to P₁ and P₂ will have normal vector ${\overline n _3}$</p> <p>Where ${\overline n _3} = \left( {{{\overline n }_1} \times {{\overline n }_2}} \right)$</p> <p>Hence, ${\overline n _3} = ( - 3, - 3,0)$</p> <p>Equation of plane E through $P(1, - 1,1)$ and ${\overline n _3}$ as normal vector</p> <p>$(x - 1,y + 1,z - 1)\,.\,( - 3, - 3,0) = 0$</p> <p>$\Rightarrow x + y = 0 \equiv E$</p> <p>Distance of $PQ(a,a,2)$ from $E = \left| {{{2a} \over {\sqrt 2 }}} \right|$</p> <p>as given, $\left| {{{2a} \over {\sqrt 2 }}} \right| = 3\sqrt 2 \Rightarrow a = \, \pm \,3$</p> <p>Hence, $Q \equiv ( \pm \,3, \pm \,3,2)$</p> <p>Distance $7Q = \sqrt {21} \Rightarrow {(PQ)^2} = 21$</p>