To solve $\log_{0.5} (x^2 - 5x + 6) \ge -1$, we must follow two conditions:

1. Domain Condition: The argument of the logarithm must be positive: $x^2 - 5x + 6 > 0 \implies (x-2)(x-3) > 0 \implies x 3$.

2. Inequality Solving: Since the base of the logarithm ($0.5$) is less than $1$, the direction of the inequality reverses when removing the logarithm: $x^2 - 5x + 6 \le (0.5)^{-1}$ $x^2 - 5x + 6 \le 2$ $x^2 - 5x + 4 \le 0 \implies (x-1)(x-4) \le 0 \implies 1 \le x \le 4$.

Now, we find the intersection of these two sets of conditions:

• Condition 1: $x \in (-\infty, 2) \cup (3, \infty)$
• Condition 2: $x \in [1, 4]$

The intersection is $x \in [1, 2) \cup (3, 4]$. Option B is incorrect because it includes the interval $[2, 3]$ where the logarithm's argument is non-positive.