1. **Stating the problem:** Solve the equation $$2 \cdot (x - 2)^2 - 3 \cdot (x - 1)^2 - (1 - x) \cdot (x - 3) = 0$$ for $x$. 2. **Formula and rules:** We will expand all terms, simplify, and solve the resulting quadratic equation. 3. **Expand each term:** $$2 \cdot (x - 2)^2 = 2 \cdot (x^2 - 4x + 4) = 2x^2 - 8x + 8$$ $$-3 \cdot (x - 1)^2 = -3 \cdot (x^2 - 2x + 1) = -3x^2 + 6x - 3$$ $$(1 - x) \cdot (x - 3) = 1 \cdot x - 1 \cdot 3 - x \cdot x + x \cdot 3 = x - 3 - x^2 + 3x = -x^2 + 4x - 3$$ 4. **Rewrite the equation:** $$2x^2 - 8x + 8 - 3x^2 + 6x - 3 - (-x^2 + 4x - 3) = 0$$ 5. **Simplify the last term carefully:** $$- (-x^2 + 4x - 3) = + x^2 - 4x + 3$$ 6. **Combine all terms:** $$2x^2 - 8x + 8 - 3x^2 + 6x - 3 + x^2 - 4x + 3 = 0$$ 7. **Group like terms:** $$ (2x^2 - 3x^2 + x^2) + (-8x + 6x - 4x) + (8 - 3 + 3) = 0$$ 8. **Calculate each group:** $$ (2 - 3 + 1)x^2 + (-8 + 6 - 4)x + (8 - 3 + 3) = 0$$ $$ 0x^2 - 6x + 8 = 0$$ 9. **Simplify:** $$-6x + 8 = 0$$ 10. **Solve for $x$:** $$-6x = -8$$ $$x = \frac{-8}{-6} = \frac{8}{6}$$ 11. **Simplify the fraction:** $$x = \frac{\cancel{8}^{4 \times 2}}{\cancel{6}^{3 \times 2}} = \frac{4}{3}$$ **Final answer:** $$\boxed{\frac{4}{3}}$$