Let P and Q be the points on the line $$\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$$ which are at a distance of 6 units from the point $$R(1, 2, 3)$$. If the centroid of the triangle PQR is $$(\alpha, \beta, \gamma)$$, then $$\alpha^2 + \beta^2 + \gamma^2$$ is:

The line is $$\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2} = t$$.

A general point on the line is $$(8t - 3, 2t + 4, 2t - 1)$$.

The distance from this point to $$R(1, 2, 3)$$ is 6:

$$(8t - 4)^2 + (2t + 2)^2 + (2t - 4)^2 = 36$$

$$64t^2 - 64t + 16 + 4t^2 + 8t + 4 + 4t^2 - 16t + 16 = 36$$

$$72t^2 - 72t + 36 = 36$$

$$72t^2 - 72t = 0$$

$$72t(t - 1) = 0$$

So $$t = 0$$ or $$t = 1$$.

For $$t = 0$$: $$P = (-3, 4, -1)$$

For $$t = 1$$: $$Q = (5, 6, 1)$$

The centroid of triangle PQR:

$$\alpha = \frac{-3 + 5 + 1}{3} = 1$$

$$\beta = \frac{4 + 6 + 2}{3} = 4$$

$$\gamma = \frac{-1 + 1 + 3}{3} = 1$$

$$\alpha^2 + \beta^2 + \gamma^2 = 1 + 16 + 1 = 18$$

The answer is $$\boxed{18}$$, which corresponds to Option (3).