1. **State the problem:** Solve the system of linear equations for $x$, $y$, and $z$: $$\begin{cases}-3x + 2y - 3z = 4 \\ x + 2y + 3z = 1 \\ 2x + 4y + 6z = 2 \end{cases}$$ 2. **Analyze the system:** Notice the third equation is exactly twice the second equation: $$2(x + 2y + 3z) = 2 \Rightarrow 2x + 4y + 6z = 2$$ This means the third equation is not independent and the system has either infinitely many solutions or no solution. 3. **Use the first two equations to solve for $x$ and $y$ in terms of $z$:** From the second equation: $$x + 2y + 3z = 1 \Rightarrow x = 1 - 2y - 3z$$ Substitute $x$ into the first equation: $$-3(1 - 2y - 3z) + 2y - 3z = 4$$ 4. **Simplify the substitution:** $$-3 + 6y + 9z + 2y - 3z = 4$$ Combine like terms: $$8y + 6z - 3 = 4$$ Add 3 to both sides: $$8y + 6z = 7$$ 5. **Solve for $y$ in terms of $z$:** $$8y = 7 - 6z \Rightarrow y = \frac{7 - 6z}{8}$$ 6. **Substitute $y$ back into the expression for $x$:** $$x = 1 - 2\left(\frac{7 - 6z}{8}\right) - 3z = 1 - \frac{2(7 - 6z)}{8} - 3z$$ Simplify the fraction: $$x = 1 - \frac{7 - 6z}{4} - 3z$$ 7. **Simplify $x$ further:** $$x = 1 - \frac{7}{4} + \frac{6z}{4} - 3z = 1 - \frac{7}{4} + \frac{3z}{2} - 3z$$ Combine constants: $$1 - \frac{7}{4} = \frac{4}{4} - \frac{7}{4} = -\frac{3}{4}$$ Combine $z$ terms: $$\frac{3z}{2} - 3z = \frac{3z}{2} - \frac{6z}{2} = -\frac{3z}{2}$$ So: $$x = -\frac{3}{4} - \frac{3z}{2}$$ 8. **Write the solution as an ordered triple:** $$\boxed{\left(-\frac{3}{4} - \frac{3}{2}z, \frac{7}{8} - \frac{3}{4}z, z\right)}$$ This represents infinitely many solutions parameterized by $z$.