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

Sum of roots: $$e_1 + e_2 = a$$ and Product of roots: $$e_1 e_2 = 2$$

For distinct real roots, the discriminant $$D > 0$$:

$$a^2 - 8 > 0 \implies a > 2\sqrt{2}$$

1. Both roots are Hyperbola eccentricities:

For a quadratic function $$f(x) = x^2 - ax + 2$$ to have both roots greater than $$1$$:

Discriminant: $$D > 0 \implies a > 2\sqrt{2}$$

Vertex position: $$-\frac{b}{2a} > 1 \implies \frac{a}{2} > 1 \implies a > 2$$

Boundary condition: $$f(1) > 0 \implies 1^2 - a(1) + 2 > 0 \implies a < 3$$

Intersecting all three conditions: $$a \in (2\sqrt{2}, 3)$$

$$\alpha = 2\sqrt{2}, \quad \beta = 3$$

2. One ellipse and one hyperbola eccentricity:

For an ellipse, $$0 < e_1 < 1$$, and for a hyperbola, $$e_2 > 1$$.

Since the product of the roots is $$e_1 e_2 = 2$$, choosing $$0 < e_1 < 1$$ automatically ensures that $$e_2 = \frac{2}{e_1} > 2$$, satisfying the hyperbola condition ($$e_2 > 1$$).

For $$f(x) = x^2 - ax + 2$$ to have exactly one root between $$0$$ and $$1$$ and the other root greater than $$1$$, the value of the function at the boundary $$x = 1$$ must be negative: $$f(1) < 0 \implies 1 - a + 2 < 0 \implies a > 3$$

$$\gamma = 3$$

$$\alpha^2 + \beta^2 + \gamma^2 = 8 + 9 + 9 = 26$$