Given the inequality $-x^2 + 3x - 2 \geq 0$,
we can rewrite it as $x^2 - 3x + 2 \leq 0$.
This can be factored into $(x - 2)(x - 1) \leq 0$.

To analyze the inequality $(x - 2)(x - 1) \leq 0$, we find the roots of the corresponding equation $(x - 2)(x - 1) = 0$, which are $x = 2$ and $x = 1$. Since the inequality is less than or equal to zero, we are interested in the interval where the product of the two factors does not exceed zero.

- For $x 2$, both factors $(x - 2)$ and $(x - 1)$ are positive, which makes their product positive.

Based on this analysis, the solution set for the inequality is the interval where the product is non-positive, which occurs when $1 \leq x \leq 2$.

Therefore, the solution set is $\boxed{\{x | 1 \leq x \leq 2\}}$.