Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2 - ax + 2 = 0$. Let the sets $\{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of hyperbolas}\} = (\alpha, \beta)$, and $\{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of an ellipse and a hyperbola, respectively}\} = (\gamma, \infty)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

The given quadratic equation is $x^2 - ax + 2 = 0$ with roots $e_1$ and $e_2$.

The sum of the roots is $e_1 + e_2 = a$ and the product is $e_1 e_2 = 2$.

For the roots to be real and distinct, the discriminant must be positive:

$\Delta = a^2 - 8 > 0 \Rightarrow a \in (-\infty, -2\sqrt{2}) \cup (2\sqrt{2}, \infty)$

Since $e_1$ and $e_2$ represent eccentricities, they must be positive. Given $e_1 e_2 = 2 > 0$, both roots have the same sign. For them to be positive, their sum must be positive, so $a > 0$. Thus, $a > 2\sqrt{2}$.

For $e_1$ and $e_2$ to be the eccentricities of hyperbolas, both must be strictly greater than $1$. This requires $1$ to lie outside the roots and the vertex of the parabola to be to the right of $1$:

$(e_1 - 1)(e_2 - 1) > 0 \Rightarrow e_1 e_2 - (e_1 + e_2) + 1 > 0$

$2 - a + 1 > 0 \Rightarrow a < 3$

Combining this with the discriminant condition, we get $a \in (2\sqrt{2}, 3)$.

Thus, $\alpha = 2\sqrt{2}$ and $\beta = 3$.

For $e_1$ and $e_2$ to be the eccentricities of an ellipse and a hyperbola, one root must be in $(0, 1)$ and the other in $(1, \infty)$. This requires $1$ to lie between the roots:

$(e_1 - 1)(e_2 - 1) < 0 \Rightarrow 2 - a + 1 < 0 \Rightarrow a > 3$

For $a > 3$, the roots are real, distinct, and positive. Thus, the set of such $a$ is $(3, \infty)$.

This gives $\gamma = 3$.

Finally, calculating the required sum:

$\alpha^2 + \beta^2 + \gamma^2 = (2\sqrt{2})^2 + 3^2 + 3^2 = 8 + 9 + 9 = 26$

Answer: $26$