Let the image of the point P(1, 2, 3) in the plane $$2x - y + z = 9$$ be Q. If the coordinates of the point R are (6, 10, 7), then the square of the area of the triangle PQR is ______.

Given: P(1,2,3), plane $$2x-y+z=9$$.

First, we find the image Q of P in the plane.

The normal to the plane is $$\vec{n} = (2,-1,1)$$.

The foot of perpendicular from P to the plane: parametric point $$(1+2t, 2-t, 3+t)$$.

Substituting in $$2x-y+z=9$$:

$$2(1+2t)-(2-t)+(3+t) = 9 \implies 3+6t = 9 \implies t = 1$$

Foot F = (3, 1, 4). Image Q = 2F - P = (5, 0, 5).

Next, we find area of triangle PQR.

R = (6, 10, 7). $$\vec{PQ} = (4, -2, 2)$$, $$\vec{PR} = (5, 8, 4)$$.

$$ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & 2 \\ 5 & 8 & 4 \end{vmatrix} = \hat{i}(-8-16) - \hat{j}(16-10) + \hat{k}(32+10) = -24\hat{i} - 6\hat{j} + 42\hat{k} $$

$$ |\vec{PQ} \times \vec{PR}|^2 = 576 + 36 + 1764 = 2376 $$

$$ \text{Area}^2 = \frac{|\vec{PQ} \times \vec{PR}|^2}{4} = \frac{2376}{4} = 594 $$

The square of the area of triangle PQR is 594.