1. **State the problem:** Simplify the expression $$\frac{1}{x-1} - \frac{3x+3}{x^2 + x - 2} + \frac{1}{x+2}$$. 2. **Factor denominators:** Note that $$x^2 + x - 2 = (x-1)(x+2)$$. 3. **Rewrite the expression with factored denominators:** $$\frac{1}{x-1} - \frac{3x+3}{(x-1)(x+2)} + \frac{1}{x+2}$$ 4. **Find common denominator:** The common denominator is $$(x-1)(x+2)$$. 5. **Rewrite each term with the common denominator:** $$\frac{1 \cdot (x+2)}{(x-1)(x+2)} - \frac{3x+3}{(x-1)(x+2)} + \frac{1 \cdot (x-1)}{(x+2)(x-1)}$$ 6. **Combine the numerators:** $$\frac{(x+2) - (3x+3) + (x-1)}{(x-1)(x+2)}$$ 7. **Simplify the numerator:** $$x + 2 - 3x - 3 + x - 1 = (x - 3x + x) + (2 - 3 - 1) = (-x) + (-2) = -x - 2$$ 8. **Rewrite the fraction:** $$\frac{-x - 2}{(x-1)(x+2)} = \frac{-(x+2)}{(x-1)(x+2)}$$ 9. **Cancel common factor $(x+2)$:** $$\frac{-\cancel{(x+2)}}{(x-1)\cancel{(x+2)}} = \frac{-1}{x-1}$$ 10. **Rewrite the final answer:** $$\frac{-1}{x-1} = \frac{1}{1-x}$$ **Final answer:** $$\boxed{\frac{1}{1-x}}$$