Let for any three distinct consecutive terms $$a, b, c$$ of an A.P, the lines $$ax + by + c = 0$$ be concurrent at the point $$P$$ and $$Q(\alpha, \beta)$$ be a point such that the system of equations $$x + y + z = 6, 2x + 5y + \alpha z = \beta$$ and $$x + 2y + 3z = 4$$, has infinitely many solutions. Then $$(PQ)^2$$ is equal to ______.

We need to find point $$P$$ where lines $$ax + by + c = 0$$ are concurrent (for $$a, b, c$$ in A.P.), and point $$Q(\alpha, \beta)$$ from the system with infinitely many solutions.

Finding point P: since $$a, b, c$$ are in A.P.: $$b = a + d$$ and $$c = a + 2d$$ for some common difference $$d$$.

Substituting into $$ax + by + c = 0$$:

$$ax + (a + d)y + (a + 2d) = 0$$

$$a(x + y + 1) + d(y + 2) = 0$$

For this to hold for all values of $$a$$ and $$d$$, we need:

$$x + y + 1 = 0$$ and $$y + 2 = 0$$

So $$y = -2$$ and $$x = 1$$. Therefore $$P = (1, -2)$$.

Finding point Q: the system is: $$x + y + z = 6$$, $$2x + 5y + \alpha z = \beta$$, $$x + 2y + 3z = 4$$.

For infinitely many solutions, the determinant of the coefficient matrix must be zero:

$$\begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix} = 1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5) = 0$$

$$15 - 2\alpha - 6 + \alpha + (-1) = 0 \implies 8 - \alpha = 0 \implies \alpha = 8$$

For consistency with $$\alpha = 8$$: From equations (1) and (3): $$(3) - (1)$$: $$y + 2z = -2$$.

From equations (1) and (2): $$(2) - 2(1)$$: $$3y + 6z = \beta - 12$$, i.e., $$3(y + 2z) = \beta - 12$$.

Since $$y + 2z = -2$$: $$3(-2) = \beta - 12 \implies \beta = 6$$.

So $$Q = (\alpha, \beta) = (8, 6)$$.

Computing $$(PQ)^2$$: $$ (PQ)^2 = (8 - 1)^2 + (6 - (-2))^2 = 49 + 64 = 113 $$

The answer is $$\boxed{113}$$.