Let a line with direction ratios $$a, -4a, -7$$ be perpendicular to the lines with direction ratios $$3, -1, 2b$$ and $$b, a, -2$$. If the point of intersection of the line $$\frac{x+1}{a^2+b^2} = \frac{y-2}{a^2-b^2} = \frac{z}{1}$$ and the plane $$x - y + z = 0$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha + \beta + \gamma$$ is equal to ________

We have a line with direction ratios $$(a, -4a, -7)$$ that is perpendicular to two lines with direction ratios $$(3, -1, 2b)$$ and $$(b, a, -2)$$. The perpendicularity conditions (dot product equals zero) give us two equations.

From the first condition: $$3a + (-4a)(-1) + (-7)(2b) = 0$$, which simplifies to $$3a + 4a - 14b = 0$$, giving $$7a = 14b$$, hence $$a = 2b$$.

From the second condition: $$a \cdot b + (-4a)(a) + (-7)(-2) = 0$$, which simplifies to $$ab - 4a^2 + 14 = 0$$. Substituting $$a = 2b$$: $$2b^2 - 4(4b^2) + 14 = 0$$, so $$2b^2 - 16b^2 + 14 = 0$$, giving $$-14b^2 = -14$$, hence $$b^2 = 1$$, and therefore $$b = \pm 1$$.

With $$b = \pm 1$$ and $$a = 2b$$, we get $$a^2 + b^2 = 4 + 1 = 5$$ and $$a^2 - b^2 = 4 - 1 = 3$$ in both cases.

The line is $$\frac{x+1}{5} = \frac{y-2}{3} = \frac{z}{1} = t$$ (say). Parametrically, $$x = -1 + 5t$$, $$y = 2 + 3t$$, $$z = t$$.

Substituting into the plane equation $$x - y + z = 0$$: $$(-1 + 5t) - (2 + 3t) + t = 0$$, which gives $$-3 + 3t = 0$$, so $$t = 1$$.

The point of intersection is $$(\alpha, \beta, \gamma) = (4, 5, 1)$$, and therefore $$\alpha + \beta + \gamma = 4 + 5 + 1 = 10$$.

Hence, the correct answer is 10.