1. **State the problem:** Solve the system of linear equations: $$\begin{cases} x + y + z = 3 \\ 13x + 2z = 2 \\ -x - 5z = -5 \end{cases}$$ 2. **Use substitution or elimination:** We have three equations with three variables $x$, $y$, and $z$. Our goal is to find values for $x$, $y$, and $z$ that satisfy all equations simultaneously. 3. **From the third equation:** $$-x - 5z = -5 \implies x + 5z = 5 \implies x = 5 - 5z$$ 4. **Substitute $x$ into the second equation:** $$13x + 2z = 2 \implies 13(5 - 5z) + 2z = 2$$ $$65 - 65z + 2z = 2$$ $$65 - 63z = 2$$ $$-63z = 2 - 65 = -63$$ $$z = \frac{-63}{-63} = 1$$ 5. **Find $x$ using $z=1$:** $$x = 5 - 5(1) = 5 - 5 = 0$$ 6. **Find $y$ using the first equation:** $$x + y + z = 3 \implies 0 + y + 1 = 3 \implies y = 3 - 1 = 2$$ 7. **Final solution:** $$\boxed{(x, y, z) = (0, 2, 1)}$$ This solution satisfies all three equations.