Solution Set of Modulus Absolute Value Equation

The given equation is of the form $|a + b| = |a| + |b|$, where $a = x$ and $b = x^2 - 9$.

This identity holds if and only if $a \cdot b \ge 0$. Therefore, we have: $x(x^2 - 9) \ge 0$

Factoring the expression gives: $x(x - 3)(x + 3) \ge 0$

Solving this inequality using the critical points $x = -3, 0, 3$, we get the solution set: $x \in [-3, 0] \cup [3, \infty)$

Comparing this with the given set $[\alpha, \beta] \cup [\gamma, \infty)$, we identify: $\alpha = -3, \quad \beta = 0, \quad \gamma = 3$

Now, calculating $\alpha^2 + \beta^2 + \gamma^2$: $\alpha^2 + \beta^2 + \gamma^2 = (-3)^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18$

Thus, the correct option is B.