1. **State the problem:** Solve the system of linear equations: $$\begin{cases} 3x + 2y + z = 2 \\ 7y + 2z = 4 \\ -10z = 6 \end{cases}$$ 2. **Start with the third equation to find $z$:** $$-10z = 6$$ Divide both sides by $-10$: $$z = \frac{6}{\cancel{-10}}\cancel{-1} = -\frac{3}{5}$$ 3. **Substitute $z = -\frac{3}{5}$ into the second equation to find $y$:** $$7y + 2\left(-\frac{3}{5}\right) = 4$$ Simplify: $$7y - \frac{6}{5} = 4$$ Add $\frac{6}{5}$ to both sides: $$7y = 4 + \frac{6}{5} = \frac{20}{5} + \frac{6}{5} = \frac{26}{5}$$ Divide both sides by 7: $$y = \frac{\frac{26}{5}}{7} = \frac{26}{5 \times 7} = \frac{26}{35}$$ 4. **Substitute $y = \frac{26}{35}$ and $z = -\frac{3}{5}$ into the first equation to find $x$:** $$3x + 2\left(\frac{26}{35}\right) + \left(-\frac{3}{5}\right) = 2$$ Simplify: $$3x + \frac{52}{35} - \frac{3}{5} = 2$$ Convert $\frac{3}{5}$ to $\frac{21}{35}$ to combine: $$3x + \frac{52}{35} - \frac{21}{35} = 2$$ $$3x + \frac{31}{35} = 2$$ Subtract $\frac{31}{35}$ from both sides: $$3x = 2 - \frac{31}{35} = \frac{70}{35} - \frac{31}{35} = \frac{39}{35}$$ Divide both sides by 3: $$x = \frac{\frac{39}{35}}{3} = \frac{39}{35 \times 3} = \frac{39}{105} = \frac{13}{35}$$ 5. **Final solution:** $$\boxed{\left(x, y, z\right) = \left(\frac{13}{35}, \frac{26}{35}, -\frac{3}{5}\right)}$$ This means the values of $x$, $y$, and $z$ satisfy all three equations simultaneously.