1. **State the problem:** Solve the system of linear equations: $$\begin{cases} 2x + y - z = 8 \\ -3x - y + 2z = -11 \\ -2x + y + 2z = -3 \end{cases}$$ 2. **Use the elimination or substitution method to solve.** Here, we will use elimination. 3. **Add the first and third equations to eliminate $x$: ** $$ (2x + y - z) + (-2x + y + 2z) = 8 + (-3) $$ $$ 0x + 2y + z = 5 $$ So, $$ 2y + z = 5 \quad (4) $$ 4. **Add the first and second equations to eliminate $y$: ** $$ (2x + y - z) + (-3x - y + 2z) = 8 + (-11) $$ $$ -x + z = -3 $$ So, $$ -x + z = -3 \Rightarrow x = z + 3 \quad (5) $$ 5. **Substitute $x = z + 3$ into the third original equation:** $$ -2(z + 3) + y + 2z = -3 $$ $$ -2z - 6 + y + 2z = -3 $$ $$ y - 6 = -3 $$ $$ y = 3 \quad (6) $$ 6. **Substitute $y=3$ into equation (4):** $$ 2(3) + z = 5 $$ $$ 6 + z = 5 $$ $$ z = -1 \quad (7) $$ 7. **Substitute $z = -1$ into equation (5):** $$ x = -1 + 3 = 2 $$ **Final solution:** $$ (x, y, z) = (2, 3, -1) $$