1. **State the problem:** Solve the system of equations by elimination: $$\begin{cases}-y + 2z = 12 \\ -x - 3y + z = 12 \\ -2x + y + 3z = -15 \end{cases}$$ 2. **Rewrite the system for clarity:** $$\begin{cases}0x - y + 2z = 12 \\ -x - 3y + z = 12 \\ -2x + y + 3z = -15 \end{cases}$$ 3. **Goal:** Eliminate variables step-by-step to solve for $x$, $y$, and $z$. 4. **Eliminate $y$ between equations (1) and (3):** Equation (1): $-y + 2z = 12$ Equation (3): $-2x + y + 3z = -15$ Add (1) and (3) to eliminate $y$: $$(-y + 2z) + (-2x + y + 3z) = 12 + (-15)$$ Simplify: $$-2x + 5z = -3$$ Label this as equation (4). 5. **Eliminate $y$ between equations (1) and (2):** Equation (1): $-y + 2z = 12$ Equation (2): $-x - 3y + z = 12$ Multiply equation (1) by 3: $$-3y + 6z = 36$$ Now subtract equation (2) from this: $$(-3y + 6z) - (-x - 3y + z) = 36 - 12$$ Simplify left side: $$-3y + 6z + x + 3y - z = x + 5z$$ Simplify right side: $$24$$ So: $$x + 5z = 24$$ Label this as equation (5). 6. **Solve equations (4) and (5):** Equation (4): $-2x + 5z = -3$ Equation (5): $x + 5z = 24$ Multiply equation (5) by 2: $$2x + 10z = 48$$ Add to equation (4): $$(-2x + 5z) + (2x + 10z) = -3 + 48$$ Simplify: $$15z = 45$$ Solve for $z$: $$z = \frac{45}{15} = 3$$ 7. **Find $x$ using equation (5):** $$x + 5(3) = 24$$ $$x + 15 = 24$$ $$x = 24 - 15 = 9$$ 8. **Find $y$ using equation (1):** $$-y + 2(3) = 12$$ $$-y + 6 = 12$$ $$-y = 6$$ $$y = -6$$ 9. **Final solution:** $$\boxed{x = 9, y = -6, z = 3}$$ This means the system's solution is $x=9$, $y=-6$, and $z=3$.