Let $$d$$ be the distance between the foot of perpendiculars of the points $$P(1, 2, -1)$$ and $$Q(2, -1, 3)$$ on the plane $$-x + y + z = 1$$. Then $$d^2$$ is equal to ______.

We need to find $$d^2$$ where $$d$$ is the distance between the feet of perpendiculars from $$P(1,2,-1)$$ and $$Q(2,-1,3)$$ to the plane $$-x + y + z = 1$$.

The normal to the plane is $$\vec{n} = (-1, 1, 1)$$. The foot of the perpendicular from $$P$$ to the plane $$-x+y+z=1$$ is $$P' = P - \frac{(-1)(1) + (1)(2) + (1)(-1) - 1}{(-1)^2 + 1^2 + 1^2} \cdot (-1, 1, 1).$$ Here the numerator is $$-1 + 2 - 1 - 1 = -1$$ and the denominator is $$1 + 1 + 1 = 3$$, giving $$P' = (1, 2, -1) - \frac{-1}{3}(-1, 1, 1) = (1, 2, -1) + \frac{1}{3}(-1, 1, 1) = \left(\frac{2}{3}, \frac{7}{3}, -\frac{2}{3}\right).$$

Similarly, for $$Q$$, we have $$Q' = Q - \frac{(-1)(2) + (1)(-1) + (1)(3) - 1}{3} \cdot (-1, 1, 1),$$ where the numerator $$-2 - 1 + 3 - 1 = -1$$, so $$Q' = (2, -1, 3) - \frac{-1}{3}(-1, 1, 1) = (2, -1, 3) + \frac{1}{3}(-1, 1, 1) = \left(\frac{5}{3}, -\frac{2}{3}, \frac{10}{3}\right).$$

The square of the distance between $$P'$$ and $$Q'$$ is $$d^2 = \left(\frac{5}{3} - \frac{2}{3}\right)^2 + \left(-\frac{2}{3} - \frac{7}{3}\right)^2 + \left(\frac{10}{3} + \frac{2}{3}\right)^2 = \left(\frac{3}{3}\right)^2 + \left(-\frac{9}{3}\right)^2 + \left(\frac{12}{3}\right)^2 = 1 + 9 + 16 = 26.$$

The answer is $$\boxed{26}$$.