Let the line $$L$$: $$\dfrac{x-1}{2} = \dfrac{y+1}{-1} = \dfrac{z-3}{1}$$ intersect the plane $$2x + y + 3z = 16$$ at the point $$P$$. Let the point $$Q$$ be the foot of perpendicular from the point $$R(1, -1, -3)$$ on the line $$L$$. If $$\alpha$$ is the area of triangle $$PQR$$, then $$\alpha^2$$ is equal to ______.

The line L is given by $$L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$$, the plane by $$2x + y + 3z = 16$$, and the point by $$R(1, -1, -3)$$. To find the intersection P of L with the plane, we express L in parametric form as $$(1+2t, -1 - t, 3 + t)$$ and substitute into the plane equation: $$2(1+2t) + (-1 - t) + 3(3 + t) = 16$$ which simplifies to $$2 + 4t - 1 - t + 9 + 3t = 16$$, i.e. $$10 + 6t = 16 \Rightarrow t = 1$$ so that $$P = (3, -2, 4)$$.

Next, to find Q, the foot of the perpendicular from R to L, note that the direction vector of L is $$\vec{d} = (2, -1, 1)$$ and a point on L is $$A = (1, -1, 3)$$. Then $$\vec{AR} = R - A = (1-1, -1+1, -3-3) = (0, 0, -6)$$, and the projection parameter is $$t = \frac{\vec{AR} \cdot \vec{d}}{|\vec{d}|^2} = \frac{0 + 0 - 6}{4 + 1 + 1} = \frac{-6}{6} = -1\,.$$ Hence $$Q = (1 + 2(-1), -1 - (-1), 3 + (-1)) = (-1, 0, 2)\,.$

Finally, for the area of triangle PQR, compute $$\vec{QP} = P - Q = (3 - (-1), -2 - 0, 4 - 2) = (4, -2, 2)$$ and $$\vec{QR} = R - Q = (1 - (-1), -1 - 0, -3 - 2) = (2, -1, -5)\,.$$ Their cross product is $$\vec{QP} \times \vec{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & 2 \\ 2 & -1 & -5 \end{vmatrix} = \hat{i}( (-2)(-5) - 2(-1)) - \hat{j}(4(-5) - 2 \cdot 2) + \hat{k}(4(-1) - (-2) \cdot 2)$$ $$= \hat{i}(10 + 2) - \hat{j}(-20 - 4) + \hat{k}(-4 + 4) = 12\hat{i} + 24\hat{j} + 0\hat{k}\,. $$ Thus $$|\vec{QP} \times \vec{QR}| = \sqrt{12^2 + 24^2} = \sqrt{144 + 576} = \sqrt{720} = 12\sqrt{5}\,, $$ and the area is $$\alpha = \frac{1}{2} \cdot 12\sqrt{5} = 6\sqrt{5}\,,\quad \alpha^2 = 180\,. $$ Therefore, the answer is $$\boxed{180}$$.