1. **State the problem:** Solve the system of linear equations: $$\begin{cases} x + 2y + z = 0 \\ 2x + y + z = 1 \\ x + y + 2z = 1 \end{cases}$$ 2. **Write the system in matrix form:** $$\begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$$ 3. **Use elimination or substitution to solve.** Start by expressing $z$ from the first equation: $$x + 2y + z = 0 \implies z = -x - 2y$$ 4. **Substitute $z$ into the second and third equations:** Second equation: $$2x + y + (-x - 2y) = 1 \implies 2x + y - x - 2y = 1 \implies (2x - x) + (y - 2y) = 1 \implies x - y = 1$$ Third equation: $$x + y + 2(-x - 2y) = 1 \implies x + y - 2x - 4y = 1 \implies (x - 2x) + (y - 4y) = 1 \implies -x - 3y = 1$$ 5. **Now solve the two-variable system:** $$\begin{cases} x - y = 1 \\ -x - 3y = 1 \end{cases}$$ Add the two equations to eliminate $x$: $$ (x - y) + (-x - 3y) = 1 + 1 \implies -4y = 2 \implies y = \frac{2}{-4} = -\frac{1}{2}$$ 6. **Find $x$ using $x - y = 1$:** $$x - \left(-\frac{1}{2}\right) = 1 \implies x + \frac{1}{2} = 1 \implies x = 1 - \frac{1}{2} = \frac{1}{2}$$ 7. **Find $z$ using $z = -x - 2y$:** $$z = -\frac{1}{2} - 2\left(-\frac{1}{2}\right) = -\frac{1}{2} + 1 = \frac{1}{2}$$ **Final solution:** $$\boxed{\left(x, y, z\right) = \left(\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}\right)}$$