1. **Stating the problem:** Solve the system of linear equations: $$\begin{cases} 2x + y + z = 3 \\ x + y + z = 1 \\ x - 2y - 3z = 4 \end{cases}$$ Find the values of $x$, $y$, and $z$ that satisfy all three equations. 2. **Method:** We will use substitution or elimination to solve the system. 3. **Step 1:** Subtract the second equation from the first to eliminate $y + z$: $$ (2x + y + z) - (x + y + z) = 3 - 1 $$ Simplify: $$ 2x - x + \cancel{y} - \cancel{y} + \cancel{z} - \cancel{z} = 2 $$ $$ x = 2 $$ 4. **Step 2:** Substitute $x=2$ into the second equation: $$ 2 + y + z = 1 $$ Rearranged: $$ y + z = 1 - 2 = -1 $$ 5. **Step 3:** Substitute $x=2$ into the third equation: $$ 2 - 2y - 3z = 4 $$ Rearranged: $$ -2y - 3z = 4 - 2 = 2 $$ 6. **Step 4:** From step 4, express $y$ in terms of $z$: $$ y = -1 - z $$ 7. **Step 5:** Substitute $y = -1 - z$ into the equation from step 3: $$ -2(-1 - z) - 3z = 2 $$ Simplify: $$ 2 + 2z - 3z = 2 $$ $$ 2 - z = 2 $$ 8. **Step 6:** Solve for $z$: $$ 2 - z = 2 $$ $$ \cancel{2} - z = \cancel{2} $$ $$ -z = 0 $$ $$ z = 0 $$ 9. **Step 7:** Substitute $z=0$ back into $y = -1 - z$: $$ y = -1 - 0 = -1 $$ 10. **Final answer:** $$ x = 2, \quad y = -1, \quad z = 0 $$ This corresponds to option D.