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 _____.

We need to find the value of $$\alpha\beta\gamma$$ for the plane $$\alpha x + \beta y + \gamma z = 1$$ passing through $$(3, -2, 5)$$ and perpendicular to the line joining $$(1, 2, 3)$$ and $$(-2, 3, 5)$$. The plane is perpendicular to the line joining $$(1,2,3)$$ and $$(-2,3,5)$$, whose direction is $$(-2-1, 3-2, 5-3) = (-3, 1, 2)$$. Thus the normal to the plane is parallel to $$(-3, 1, 2)$$, so $$(\alpha, \beta, \gamma) = k(-3, 1, 2)$$ for some constant $$k$$.

The plane passes through $$(3, -2, 5)$$, so $$\alpha(3) + \beta(-2) + \gamma(5) = 1$$, giving $$k(-9 - 2 + 10) = 1$$, hence $$k(-1) = 1$$ and $$k = -1$$. Therefore, $$\alpha = 3, \beta = -1, \gamma = -2$$, and so $$\alpha\beta\gamma = 3 \times (-1) \times (-2) = 6$$.

Therefore, the answer is 6.