1. **State the problem:** Solve the system of linear equations: $$\begin{cases} x + y + z = 6 \\ 2x - y + 3z = 9 \\ -x + 2y + 2z = 9 \end{cases}$$ 2. **Use substitution or elimination method.** Here, we use elimination. 3. From the first equation, express $x$: $$x = 6 - y - z$$ 4. Substitute $x$ into the second and third equations: $$2(6 - y - z) - y + 3z = 9$$ $$-(6 - y - z) + 2y + 2z = 9$$ 5. Simplify the second equation: $$12 - 2y - 2z - y + 3z = 9$$ $$12 - 3y + z = 9$$ 6. Simplify the third equation: $$-6 + y + z + 2y + 2z = 9$$ $$-6 + 3y + 3z = 9$$ 7. Rearrange both: $$-3y + z = 9 - 12$$ $$-3y + z = -3$$ $$3y + 3z = 9 + 6$$ $$3y + 3z = 15$$ 8. Simplify the second: $$y + z = 5$$ 9. Now solve the system: $$\begin{cases} -3y + z = -3 \\ y + z = 5 \end{cases}$$ 10. Subtract the first from the second: $$\cancel{y} + z - (-3y + z) = 5 - (-3)$$ $$y + z + 3y - z = 5 + 3$$ $$4y = 8$$ 11. Solve for $y$: $$y = \frac{8}{4} = 2$$ 12. Substitute $y=2$ into $y + z = 5$: $$2 + z = 5$$ $$z = 3$$ 13. Substitute $y=2$ and $z=3$ into $x = 6 - y - z$: $$x = 6 - 2 - 3 = 1$$ **Final answer:** $$\boxed{(x, y, z) = (1, 2, 3)}$$