1. **Stating the problem:** Solve the system of linear equations for $x_3$: $$\begin{cases} 2x_1 - x_2 - x_3 = 4 \\ 3x_1 + 4x_2 - 2x_3 = 11 \\ 3x_1 - 2x_2 + 4x_3 = 11 \end{cases}$$ 2. **Method:** We will use substitution or elimination to find $x_3$. First, express $x_3$ from the first equation: $$2x_1 - x_2 - x_3 = 4 \implies x_3 = 2x_1 - x_2 - 4$$ 3. **Substitute $x_3$ into the second and third equations:** Second equation: $$3x_1 + 4x_2 - 2(2x_1 - x_2 - 4) = 11$$ Simplify: $$3x_1 + 4x_2 - 4x_1 + 2x_2 + 8 = 11$$ $$(-x_1) + 6x_2 + 8 = 11$$ $$-x_1 + 6x_2 = 3$$ Third equation: $$3x_1 - 2x_2 + 4(2x_1 - x_2 - 4) = 11$$ Simplify: $$3x_1 - 2x_2 + 8x_1 - 4x_2 - 16 = 11$$ $$11x_1 - 6x_2 - 16 = 11$$ $$11x_1 - 6x_2 = 27$$ 4. **Now solve the system:** $$\begin{cases} -x_1 + 6x_2 = 3 \\ 11x_1 - 6x_2 = 27 \end{cases}$$ Add the two equations to eliminate $x_2$: $$(-x_1 + 6x_2) + (11x_1 - 6x_2) = 3 + 27$$ $$10x_1 = 30 \implies x_1 = 3$$ 5. **Find $x_2$ using $-x_1 + 6x_2 = 3$:** $$-3 + 6x_2 = 3 \implies 6x_2 = 6 \implies x_2 = 1$$ 6. **Find $x_3$ using $x_3 = 2x_1 - x_2 - 4$:** $$x_3 = 2(3) - 1 - 4 = 6 - 1 - 4 = 1$$ **Final answer:** $$\boxed{x_3 = 1}$$