1. **Problem Statement:** Solve the system of linear equations: $$\begin{cases} 3x + 3y + 2z = 1 \\ x + 2y = 4 \\ 10y + 3z = -2 \\ 2x - 3y - z = 5 \end{cases}$$ 2. **Step 1: Analyze the system** We have 4 equations but only 3 variables ($x,y,z$). This is an overdetermined system, so we check for consistency. 3. **Step 2: Use equations with 3 variables first** From equation 2: $x + 2y = 4 \Rightarrow x = 4 - 2y$ 4. **Step 3: Substitute $x$ into other equations** Equation 1: $3(4 - 2y) + 3y + 2z = 1 \Rightarrow 12 - 6y + 3y + 2z = 1 \Rightarrow 12 - 3y + 2z = 1$ Simplify: $$-3y + 2z = 1 - 12 = -11$$ Equation 4: $2(4 - 2y) - 3y - z = 5 \Rightarrow 8 - 4y - 3y - z = 5 \Rightarrow 8 - 7y - z = 5$ Simplify: $$-7y - z = 5 - 8 = -3$$ 5. **Step 4: Use equation 3:** $$10y + 3z = -2$$ 6. **Step 5: Solve the system of 3 equations with variables $y$ and $z$:** $$\begin{cases} -3y + 2z = -11 \\ 10y + 3z = -2 \\ -7y - z = -3 \end{cases}$$ 7. **Step 6: Solve first two equations for $y$ and $z$:** Multiply first equation by 3: $$-9y + 6z = -33$$ Multiply second equation by 2: $$20y + 6z = -4$$ Subtract first from second: $$(20y + 6z) - (-9y + 6z) = -4 - (-33) \Rightarrow 20y + 6z + 9y - 6z = 29 \Rightarrow 29y = 29 \Rightarrow y = 1$$ 8. **Step 7: Substitute $y=1$ into $-3y + 2z = -11$:** $$-3(1) + 2z = -11 \Rightarrow -3 + 2z = -11 \Rightarrow 2z = -8 \Rightarrow z = -4$$ 9. **Step 8: Substitute $y=1$ and $z=-4$ into $x = 4 - 2y$:** $$x = 4 - 2(1) = 4 - 2 = 2$$ 10. **Step 9: Verify with equation 4:** $$2x - 3y - z = 2(2) - 3(1) - (-4) = 4 - 3 + 4 = 5$$ Correct. **Final solution:** $$\boxed{x=2, y=1, z=-4}$$