Let $$\alpha x + \beta y + \gamma z = 1$$ be the equation of a plane passing through the point $$(3, -2, 5)$$ and perpendicular to the line joining the points $$(1, 2, 3)$$ and $$(-2, 3, 5)$$. Then the value of $$\alpha \beta y$$ is equal to ______.

Find the plane $$\alpha x + \beta y + \gamma z = 1$$ through $$(3, -2, 5)$$ perpendicular to the line joining $$(1, 2, 3)$$ and $$(-2, 3, 5)$$.

$$\vec{n} = (-2 - 1,\ 3 - 2,\ 5 - 3) = (-3, 1, 2)$$

Normal $$(- 3, 1, 2)$$ and passing through $$(3, -2, 5)$$:

$$-3(x - 3) + 1(y + 2) + 2(z - 5) = 0$$

$$-3x + 9 + y + 2 + 2z - 10 = 0$$

$$-3x + y + 2z + 1 = 0 \implies 3x - y - 2z = 1$$

Comparing with $$\alpha x + \beta y + \gamma z = 1$$:

$$\alpha = 3,\ \beta = -1,\ \gamma = -2$$

$$\alpha\beta\gamma = 3 \times (-1) \times (-2) = 6$$

The answer is $$\boxed{6}$$.