Let the line $$\dfrac{x-3}{7} = \dfrac{y-2}{-1} = \dfrac{z-3}{-4}$$ intersect the plane containing the lines $$\dfrac{x-4}{1} = \dfrac{y+1}{-2} = \dfrac{z}{1}$$ and $$4ax - y + 5z - 7a = 0 = 2x - 5y - z - 3$$, $$a \in \mathbb{R}$$ at the point $$P(\alpha, \beta, \gamma)$$. Then the value of $$\alpha + \beta + \gamma$$  equals ______.

The line of intersection of $$4ax - y + 5z - 7a = 0$$ and $$2x - 5y - z - 3 = 0$$ lies in every plane of the family $$(4ax - y + 5z - 7a) + \mu(2x - 5y - z - 3) = 0$$, i.e., $$(4a + 2\mu)x - (1 + 5\mu)y + (5 - \mu)z - (7a + 3\mu) = 0$$ $$-(1)$$.

The plane must contain the line $$\dfrac{x-4}{1} = \dfrac{y+1}{-2} = \dfrac{z}{1}$$, which passes through $$(4, -1, 0)$$ with direction $$(1, -2, 1)$$.

Substituting $$(4, -1, 0)$$ into $$(1)$$: $$4(4a + 2\mu) + (1 + 5\mu) + 0 = 7a + 3\mu$$, giving $$16a + 8\mu + 1 + 5\mu = 7a + 3\mu$$, so $$9a + 10\mu + 1 = 0$$ $$-(2)$$.

The direction $$(1, -2, 1)$$ must be perpendicular to the plane normal: $$(4a + 2\mu) - 2(-(1 + 5\mu)) + (5 - \mu) = 0$$, giving $$4a + 2\mu + 2 + 10\mu + 5 - \mu = 0$$, so $$4a + 11\mu + 7 = 0$$ $$-(3)$$.

From $$(2)$$: $$a = \dfrac{-10\mu - 1}{9}$$. Substituting into $$(3)$$: $$\dfrac{4(-10\mu - 1)}{9} + 11\mu + 7 = 0$$, giving $$-40\mu - 4 + 99\mu + 63 = 0$$, so $$59\mu = -59$$, thus $$\mu = -1$$.

From $$(2)$$: $$9a - 10 + 1 = 0$$, so $$a = 1$$.

The plane equation becomes $$(4-2)x - (1-5)y + (5+1)z - (7-3) = 0$$, i.e., $$2x + 4y + 6z - 4 = 0$$, simplified to $$x + 2y + 3z = 2$$.

The line $$\dfrac{x-3}{7} = \dfrac{y-2}{-1} = \dfrac{z-3}{-4} = t$$ gives points $$(3+7t, 2-t, 3-4t)$$. Substituting into the plane:

$$(3+7t) + 2(2-t) + 3(3-4t) = 2$$, so $$3 + 7t + 4 - 2t + 9 - 12t = 2$$, giving $$16 - 7t = 2$$, thus $$t = 2$$.

The intersection point is $$P = (17, 0, -5)$$, and $$\alpha + \beta + \gamma = 17 + 0 - 5 = 12$$.

The answer is $$12$$.