1. **Stating the problem:** Solve the system of equations: $$\begin{cases} x + 3y - z = -2 \\ 3x + 6y - 3z = -5 \\ 2x + 4y + z = \frac{11}{3} \end{cases}$$ 2. **Analyze the system:** Notice the second equation is a multiple of the first equation if consistent. Let's check: Multiply the first equation by 3: $$3(x + 3y - z) = 3(-2) \Rightarrow 3x + 9y - 3z = -6$$ But the second equation is: $$3x + 6y - 3z = -5$$ Since $9y \neq 6y$ and $-6 \neq -5$, the second equation is not a multiple of the first, so the system is consistent and independent. 3. **Use substitution or elimination:** Let's use elimination. From the first equation: $$x = -2 - 3y + z$$ Substitute $x$ into the third equation: $$2(-2 - 3y + z) + 4y + z = \frac{11}{3}$$ Simplify: $$-4 - 6y + 2z + 4y + z = \frac{11}{3}$$ $$-4 - 2y + 3z = \frac{11}{3}$$ Add 4 to both sides: $$-2y + 3z = \frac{11}{3} + 4 = \frac{11}{3} + \frac{12}{3} = \frac{23}{3}$$ 4. **Express $y$ in terms of $z$:** $$-2y = \frac{23}{3} - 3z \Rightarrow y = \frac{3z - \frac{23}{3}}{2} = \frac{3z}{2} - \frac{23}{6}$$ 5. **Substitute $x$ and $y$ into the second equation:** Second equation: $$3x + 6y - 3z = -5$$ Substitute $x = -2 - 3y + z$: $$3(-2 - 3y + z) + 6y - 3z = -5$$ Simplify: $$-6 - 9y + 3z + 6y - 3z = -5$$ $$-6 - 3y = -5$$ Add 6 to both sides: $$-3y = 1 \Rightarrow y = -\frac{1}{3}$$ 6. **Find $z$ using $y$ expression:** Recall: $$y = \frac{3z}{2} - \frac{23}{6} = -\frac{1}{3}$$ Multiply both sides by 6: $$9z - 23 = -2$$ Add 23: $$9z = 21 \Rightarrow z = \frac{21}{9} = \frac{7}{3}$$ 7. **Find $x$ using $x = -2 - 3y + z$:** $$x = -2 - 3\left(-\frac{1}{3}\right) + \frac{7}{3} = -2 + 1 + \frac{7}{3} = -1 + \frac{7}{3} = \frac{-3}{3} + \frac{7}{3} = \frac{4}{3}$$ **Final answer:** $$\boxed{\left(x, y, z\right) = \left(\frac{4}{3}, -\frac{1}{3}, \frac{7}{3}\right)}$$