1. **Stating the problem:** Solve the system of linear equations for part (a): $$\begin{cases} 2x + y - 3z = 3 \\ x - y + 5z = -2 \\ 2x + 2y - z = -1 \end{cases}$$ 2. **Isolate $y$ from the first equation:** $$2x + y - 3z = 3 \implies y = -2x + 3z + 3$$ 3. **Substitute $y$ into the second equation:** $$x - y + 5z = -2$$ Replace $y$: $$x - (-2x + 3z + 3) + 5z = -2$$ Simplify: $$x + 2x - 3z - 3 + 5z = -2$$ $$3x + 2z - 3 = -2$$ Add 3 to both sides: $$3x + 2z = 1$$ 4. **Substitute $y$ into the third equation:** $$2x + 2y - z = -1$$ Replace $y$: $$2x + 2(-2x + 3z + 3) - z = -1$$ Simplify: $$2x - 4x + 6z + 6 - z = -1$$ $$-2x + 5z + 6 = -1$$ Subtract 6 from both sides: $$-2x + 5z = -7$$ 5. **Solve the system of two equations with two variables:** $$\begin{cases} 3x + 2z = 1 \\ -2x + 5z = -7 \end{cases}$$ Multiply the first equation by 2 and the second by 3 to eliminate $x$: $$\begin{cases} 6x + 4z = 2 \\ -6x + 15z = -21 \end{cases}$$ Add the two equations: $$6x - 6x + 4z + 15z = 2 - 21$$ $$19z = -19$$ Divide both sides by 19: $$z = \frac{\cancel{19}z}{\cancel{19}} = \frac{-19}{19} = -1$$ 6. **Find $x$ by substituting $z = -1$ into $3x + 2z = 1$:** $$3x + 2(-1) = 1$$ $$3x - 2 = 1$$ Add 2 to both sides: $$3x = 3$$ Divide both sides by 3: $$x = \frac{\cancel{3}x}{\cancel{3}} = \frac{3}{3} = 1$$ 7. **Find $y$ by substituting $x=1$ and $z=-1$ into $y = -2x + 3z + 3$:** $$y = -2(1) + 3(-1) + 3 = -2 - 3 + 3 = -2$$ **Final solution:** $$\boxed{(x, y, z) = (1, -2, -1)}$$