Given, plane $P_1=2x-2y+z=0$,

Whose normal vector is $\equiv {\overline{n}}_1=\left(2,-2,1\right)$

Second plane, $P_2\equiv x-y+2z=4$,

Whose normal vector is $\equiv {\overline{n}}_2=\left(1,-1,2\right)$

Now let plane perpendicular to $P_1$ and $P_2$ will have normal vector ${\overline{n}}_3$

Where ${\overline{n}}_3=\left({\overline{n}}_1\times {\overline{n}}_2\right)$

$\Rightarrow {\overrightarrow{n}}_3=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 1\\ 1& -1& 2\end{array}\right|=-3\hat{i}-3\hat{j}$

Hence, ${\overline{n}}_3=\left(-3,-3,0\right)$

Equation of plane $E$ through $P\left(1,-1,1\right)$ and ${\overline{n}}_3$ as normal vector will be,

$-3\left(x-1\right)-3\left(y+1\right)+0\left(z-1\right)=0$

$\Rightarrow x+y=0\equiv E$

Now distance of point $Q\left(a,a,2\right)$ from $E=\left|\frac{2a}{\sqrt{2}}\right|$

Now given, $\left|\frac{2a}{\sqrt{2}}\right|=3\sqrt{2}\Rightarrow a=\pm 3$

Hence, $Q\equiv \left(\pm 3,\pm 3,2\right)$

Now distance $PQ=\sqrt{21}\Rightarrow {\left(PQ\right)}^2=21$