If $$\alpha$$ satisfies the equation $$x^2 + x + 1 = 0$$ and $$(1 + \alpha)^7 = A + B\alpha + C\alpha^2$$, $$A, B, C \geq 0$$, then $$5(3A - 2B - C)$$ is equal to _______.

$$\alpha$$ satisfies $$x^2 + x + 1 = 0$$, so $$\alpha$$ is a primitive cube root of unity ($$\omega$$).

We know $$\alpha^2 + \alpha + 1 = 0$$, so $$\alpha^2 = -\alpha - 1$$ and $$\alpha^3 = 1$$.

$$(1 + \alpha)^7$$: Since $$1 + \alpha + \alpha^2 = 0$$, we have $$1 + \alpha = -\alpha^2$$.

$$(1 + \alpha)^7 = (-\alpha^2)^7 = -\alpha^{14} = -(\alpha^3)^4 \cdot \alpha^2 = -1 \cdot \alpha^2 = -\alpha^2$$

Since $$\alpha^2 = -\alpha - 1$$:

$$-\alpha^2 = \alpha + 1$$

So $$(1 + \alpha)^7 = 1 + \alpha = A + B\alpha + C\alpha^2$$.

Comparing: $$A = 1$$, $$B = 1$$, $$C = 0$$.

$$5(3A - 2B - C) = 5(3 - 2 - 0) = 5$$

The answer is $$\boxed{5}$$.