1. **State the problem:** Solve the system of linear equations: $$\begin{cases} 2x + 6y - z = 6 \\ 4x - 3y + 5z = -5 \\ 6x + 9y - 2z = 11 \end{cases}$$ 2. **Use elimination or substitution to solve.** We will use elimination here. 3. Multiply the first equation by 2 to help eliminate $z$ with the third equation: $$2(2x + 6y - z) = 2(6) \Rightarrow 4x + 12y - 2z = 12$$ 4. Subtract the third equation from this result: $$\cancel{4x} + 12y - 2z - (\cancel{6x} + 9y - 2z) = 12 - 11$$ $$4x + 12y - 2z - 6x - 9y + 2z = 1$$ $$-2x + 3y = 1$$ 5. Now we have a simpler equation: $$-2x + 3y = 1$$ 6. Next, multiply the first original equation by 5 to help eliminate $z$ with the second equation: $$5(2x + 6y - z) = 5(6) \Rightarrow 10x + 30y - 5z = 30$$ 7. Add this to the second equation: $$10x + 30y - 5z + (4x - 3y + 5z) = 30 + (-5)$$ $$10x + 30y - 5z + 4x - 3y + 5z = 25$$ $$14x + 27y = 25$$ 8. Now solve the system: $$\begin{cases} -2x + 3y = 1 \\ 14x + 27y = 25 \end{cases}$$ 9. Multiply the first equation by 7: $$7(-2x + 3y) = 7(1) \Rightarrow -14x + 21y = 7$$ 10. Add this to the second equation: $$-14x + 21y + 14x + 27y = 7 + 25$$ $$48y = 32$$ 11. Solve for $y$: $$y = \frac{32}{48} = \frac{2}{3}$$ 12. Substitute $y=\frac{2}{3}$ into $-2x + 3y = 1$: $$-2x + 3\left(\frac{2}{3}\right) = 1$$ $$-2x + 2 = 1$$ 13. Solve for $x$: $$-2x = 1 - 2 = -1$$ $$x = \frac{-1}{-2} = \frac{1}{2}$$ 14. Substitute $x=\frac{1}{2}$ and $y=\frac{2}{3}$ into the first original equation to find $z$: $$2\left(\frac{1}{2}\right) + 6\left(\frac{2}{3}\right) - z = 6$$ $$1 + 4 - z = 6$$ $$5 - z = 6$$ 15. Solve for $z$: $$-z = 6 - 5 = 1$$ $$z = -1$$ **Final answer:** $$(x, y, z) = \left(\frac{1}{2}, \frac{2}{3}, -1\right)$$