If the range of the function $$f(x) = \frac{5 - x}{x^2 - 3x + 2}$$, $$x \ne 1, 2$$, is $$(-\infty, \alpha] \cup [\beta, \infty)$$, then $$\alpha^2 + \beta^2$$ is equal to :

Let the given function be $$y = f(x) = \frac{5 - x}{x^{2} - 3x + 2}$$, where $$x \neq 1,\,2$$ because the denominator $$x^{2} - 3x + 2 = (x - 1)(x - 2)$$ vanishes at these points.

To find the range we eliminate $$x$$.
Cross-multiply:

$$y(x^{2} - 3x + 2) = 5 - x$$

Rearrange into a quadratic in $$x$$:

$$y x^{2} - 3y x + 2y + x - 5 = 0$$

Group like terms:

$$y x^{2} + (-3y + 1)x + (2y - 5) = 0 \quad -(1)$$

For a given real $$y$$ to lie in the range, equation $$(1)$$ must have at least one real root $$x$$ that is different from $$1$$ and $$2$$. The first requirement is that its discriminant is non-negative.

Discriminant $$\Delta$$ of $$(1)$$:

$$\Delta = (-3y + 1)^{2} - 4y(2y - 5)$$

Simplify:

$$\Delta = 9y^{2} - 6y + 1 - 8y^{2} + 20y = y^{2} + 14y + 1$$

Thus real roots exist when

$$y^{2} + 14y + 1 \ge 0 \quad -(2)$$

Factor $$(2)$$ by finding its roots. Solve

$$y^{2} + 14y + 1 = 0$$

Roots:

$$y = \frac{-14 \pm \sqrt{14^{2} - 4 \cdot 1 \cdot 1}}{2} = \frac{-14 \pm \sqrt{196 - 4}}{2} = \frac{-14 \pm \sqrt{192}}{2} = -7 \pm 4\sqrt{3}$$

Denote

$$\alpha = -7 - 4\sqrt{3}, \qquad \beta = -7 + 4\sqrt{3}$$

Since the coefficient of $$y^{2}$$ in $$(2)$$ is positive, the inequality $$\Delta \ge 0$$ holds outside the interval formed by these roots:

$$y \le \alpha \quad \text{or} \quad y \ge \beta$$

Hence the range is $$(-\infty,\,\alpha] \cup [\beta,\,\infty)$$ as stated.

Now compute $$\alpha^{2} + \beta^{2}$$.

For the quadratic $$y^{2} + 14y + 1 = 0$$:
  Sum of roots $$\alpha + \beta = -14$$
  Product of roots $$\alpha\beta = 1$$

Use the identity $$\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta$$:

$$\alpha^{2} + \beta^{2} = (-14)^{2} - 2 \cdot 1 = 196 - 2 = 194$$

Therefore, $$\alpha^{2} + \beta^{2} = 194$$.

Answer: Option D