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 $$3\sqrt{2}$$, then $$(PQ)^2$$ is equal to

Plane $$E$$ is perpendicular to the planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4$$ and passes through $$P(1, -1, 1)$$, so the normal vectors of the two given planes are $$\vec{n}_1 = (2, -2, 1)$$ and $$\vec{n}_2 = (1, -1, 2)$$ and the normal to plane $$E$$ is parallel to their cross product:

$$ \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -1 & 2 \end{vmatrix} = (-4+1)\hat{i} - (4-1)\hat{j} + (-2+2)\hat{k} = (-3, -3, 0) $$

Thus a direction for the normal is $$(1, 1, 0)$$. Using this normal, the equation of plane $$E$$ through $$P(1, -1, 1)$$ is

$$ 1(x - 1) + 1(y + 1) + 0(z - 1) = 0 \implies x + y = 0 $$

The distance from $$Q(a, a, 2)$$ to the plane $$x + y = 0$$ is

$$ d = \frac{|a + a|}{\sqrt{1^2 + 1^2}} = \frac{|2a|}{\sqrt{2}} = |a|\sqrt{2} = 3\sqrt{2} $$

Hence $$|a| = 3$$, so $$a = 3$$ or $$a = -3$$.

Next, compute $$(PQ)^2$$:

$$ (PQ)^2 = (a-1)^2 + (a+1)^2 + (2-1)^2 = 2a^2 + 2 + 1 = 2a^2 + 3 $$

Since $$a^2 = 9$$,

$$ (PQ)^2 = 18 + 3 = 21 $$

The answer is Option C: $$21$$.