1. **State the problem:** Solve the system of equations: $$\begin{cases} 2x + 3y - z = 5 \\ x - y + 4z = 2 \\ 3x + 2y + z = 7 \end{cases}$$ 2. **Method:** We will use the substitution or elimination method to find $x$, $y$, and $z$. 3. From the second equation, express $x$ in terms of $y$ and $z$: $$x = y - 4z + 2$$ 4. Substitute $x$ into the first and third equations: - First equation: $$2(y - 4z + 2) + 3y - z = 5$$ Simplify: $$2y - 8z + 4 + 3y - z = 5$$ $$5y - 9z + 4 = 5$$ $$5y - 9z = 1$$ - Third equation: $$3(y - 4z + 2) + 2y + z = 7$$ Simplify: $$3y - 12z + 6 + 2y + z = 7$$ $$5y - 11z + 6 = 7$$ $$5y - 11z = 1$$ 5. Now solve the system: $$\begin{cases} 5y - 9z = 1 \\ 5y - 11z = 1 \end{cases}$$ 6. Subtract the second equation from the first: $$(5y - 9z) - (5y - 11z) = 1 - 1$$ $$5y - 9z - 5y + 11z = 0$$ $$2z = 0$$ $$z = 0$$ 7. Substitute $z=0$ into $5y - 9z = 1$: $$5y - 0 = 1$$ $$5y = 1$$ $$y = \frac{1}{5}$$ 8. Substitute $y=\frac{1}{5}$ and $z=0$ into $x = y - 4z + 2$: $$x = \frac{1}{5} - 0 + 2 = \frac{1}{5} + 2 = \frac{1}{5} + \frac{10}{5} = \frac{11}{5}$$ **Final answer:** $$\boxed{\left(x, y, z\right) = \left(\frac{11}{5}, \frac{1}{5}, 0\right)}$$