If the set of all solutions of $|x^2 + x - 9| = |x| + |x^2 - 9|$ is $[\alpha, \beta] \cup [\gamma, \infty)$, then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to:

The given equation is $|x^2 + x - 9| = |x| + |x^2 - 9|$.

Let $A = x$ and $B = x^2 - 9$. Then $A + B = x^2 + x - 9$.

The equation is of the form $|A + B| = |A| + |B|$, which holds true if and only if $A \cdot B \ge 0$.

Substituting the values of $A$ and $B$:

$x(x^2 - 9) \ge 0$

$x(x - 3)(x + 3) \ge 0$

The critical points are $-3, 0, 3$. Using the wavy curve method, the solution set is:

$x \in [-3, 0] \cup [3, \infty)$

Comparing this with the given solution set $[\alpha, \beta] \cup [\gamma, \infty)$, we get:

$\alpha = -3$, $\beta = 0$, $\gamma = 3$

We need to find the value of $\alpha^2 + \beta^2 + \gamma^2$:

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

Answer: $18$