The factor (x - 1)² is always non-negative for all real x. Therefore, the sign of the expression depends on (x - 2) / (x - 3), except at x = 1 where the expression is 0 (which satisfies ≥ 0). Solving (x - 2) / (x - 3) ≥ 0: critical points are 2 and 3. Testing intervals gives (-∞, 2] ∪ (3, ∞). Since x = 1 is already included in the interval (-∞, 2], the final solution is x ∈ (-∞, 2] ∪ (3, ∞).