The direction vector of the line is $$\vec d = (2, -1, 2)$$, and its magnitude is:
$$|\vec d| = \sqrt{2^2 + (-1)^2 + 2^2} = 3$$
Given that the base $$BC = 6$$ and the $$\text{Area of } \triangle ABC = 3\sqrt{10}$$, the perpendicular height $$h$$ from point $$A$$ to the line is:
$$h = \frac{2 \times \text{Area}}{BC} = \frac{2 \times 3\sqrt{10}}{6} = \sqrt{10}$$
Let a fixed point on the line be $$P(\alpha+1, 1, 3)$$.
Given $$A(\alpha, 2, \alpha)$$, the vector $$\overrightarrow{PA}$$ is:
$$\overrightarrow{PA} = (\alpha - (\alpha+1))\hat i + (2 - 1)\hat j + (\alpha - 3)\hat k = -\hat i + \hat j + (\alpha - 3)\hat k$$
Now, find the cross product $$\overrightarrow{PA} \times \vec d$$:
$$\overrightarrow{PA} \times \vec d = \begin{vmatrix}\hat i & \hat j & \hat k \\ -1 & 1 & \alpha-3 \\ 2 & -1 & 2 \end{vmatrix}$$ $$= \big(2 - (-1)(\alpha-3)\big)\hat i - \big(-2 - 2(\alpha-3)\big)\hat j + \big(1 - 2\big)\hat k$$ $$= (\alpha - 1)\hat i - (4 - 2\alpha)\hat j - \hat k$$
The perpendicular distance $$h$$ is also given by the formula $$h = \frac{|\overrightarrow{PA} \times \vec d|}{|\vec d|}$$. Therefore:
$$\frac{\sqrt{(\alpha - 1)^2 + (4 - 2\alpha)^2 + (-1)^2}}{3} = \sqrt{10}$$
Squaring both sides:
$$(\alpha - 1)^2 + (4 - 2\alpha)^2 + 1 = 90$$
Expand the terms:
$$(\alpha^2 - 2\alpha + 1) + (16 - 16\alpha + 4\alpha^2) + 1 = 90$$ $$5\alpha^2 - 18\alpha + 18 = 90$$ $$5\alpha^2 - 18\alpha - 72 = 0$$
Factor the quadratic equation:
$$5\alpha^2 - 30\alpha + 12\alpha - 72 = 0$$ $$5\alpha(\alpha - 6) + 12(\alpha - 6) = 0$$ $$(5\alpha + 12)(\alpha - 6) = 0$$
Since $$\alpha \in \mathbb{Z}$$ (it must be an integer), we select:
$$\alpha = 6$$
$$\alpha^2 = 6^2 = 36$$
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