Let a line $$L$$ pass through the point $$P(2, 3, 1)$$ and be parallel to the line $$x + 3y - 2z - 2 = 0 = x - y + 2z$$. If the distance of $$L$$ from the point $$(5, 3, 8)$$ is $$\alpha$$, then $$3\alpha^2$$ is equal to ______.

Line $$L$$ passes through $$P(2, 3, 1)$$ and is parallel to the line of intersection of the planes $$x + 3y - 2z - 2 = 0$$ and $$x - y + 2z = 0$$.

Now, find the direction of $$L$$.

The direction is given by the cross product of the normals of the two planes:

$$\vec{n_1} = (1, 3, -2)$$, $$\vec{n_2} = (1, -1, 2)$$

$$ \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ 1 & -1 & 2 \end{vmatrix} = \hat{i}(6-2) - \hat{j}(2+2) + \hat{k}(-1-3) = (4, -4, -4) $$

Direction vector: $$(1, -1, -1)$$

Now, find the distance from $$(5, 3, 8)$$ to line $$L$$.

Let $$\vec{PQ} = (5-2, 3-3, 8-1) = (3, 0, 7)$$ and $$\vec{d} = (1, -1, -1)$$.

The distance is given by:

$$ \alpha = \frac{|\vec{PQ} \times \vec{d}|}{|\vec{d}|} $$

$$ \vec{PQ} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & 7 \\ 1 & -1 & -1 \end{vmatrix} = \hat{i}(0+7) - \hat{j}(-3-7) + \hat{k}(-3-0) = (7, 10, -3) $$

$$ |\vec{PQ} \times \vec{d}| = \sqrt{49 + 100 + 9} = \sqrt{158} $$

$$ |\vec{d}| = \sqrt{1 + 1 + 1} = \sqrt{3} $$

$$ \alpha = \frac{\sqrt{158}}{\sqrt{3}} $$

$$ \alpha^2 = \frac{158}{3} $$

$$ 3\alpha^2 = 158 $$