1. **State the problem:** Solve the system of linear equations: $$\begin{cases} 3x + y + z = 5 \\ -2x - y - 3z = -6 \\ 5x - y + 11z = 15 \end{cases}$$ 2. **Use the elimination or substitution method to solve for variables.** Here, we will use elimination. 3. Add the first and second equations to eliminate $y$: $$ (3x + y + z) + (-2x - y - 3z) = 5 + (-6) $$ $$ 3x - 2x + y - y + z - 3z = -1 $$ $$ x - 2z = -1 $$ 4. Add the first and third equations to eliminate $y$: $$ (3x + y + z) + (5x - y + 11z) = 5 + 15 $$ $$ 3x + 5x + y - y + z + 11z = 20 $$ $$ 8x + 12z = 20 $$ 5. Simplify the second equation: $$ 8x + 12z = 20 $$ Divide both sides by 4: $$ \cancel{4} \times 2x + \cancel{4} \times 3z = \cancel{4} \times 5 $$ $$ 2x + 3z = 5 $$ 6. Now solve the system: $$ \begin{cases} x - 2z = -1 \\ 2x + 3z = 5 \end{cases} $$ 7. Multiply the first equation by 2: $$ 2x - 4z = -2 $$ 8. Subtract this from the second equation: $$ (2x + 3z) - (2x - 4z) = 5 - (-2) $$ $$ 2x + 3z - 2x + 4z = 7 $$ $$ 7z = 7 $$ 9. Solve for $z$: $$ z = \frac{7}{7} = 1 $$ 10. Substitute $z=1$ into $x - 2z = -1$: $$ x - 2(1) = -1 $$ $$ x - 2 = -1 $$ $$ x = -1 + 2 = 1 $$ 11. Substitute $x=1$ and $z=1$ into the first original equation to find $y$: $$ 3(1) + y + 1 = 5 $$ $$ 3 + y + 1 = 5 $$ $$ y + 4 = 5 $$ $$ y = 5 - 4 = 1 $$ **Final answer:** $$ (x, y, z) = (1, 1, 1) $$