1. **State the problem:** Solve the system of equations: $$\begin{cases} x - 3y - 3z = 7 \\ 2x + y + x = 7 \\ x + 2y - 2z = -6 \end{cases}$$ 2. **Simplify the second equation:** $$2x + y + x = 7 \implies 3x + y = 7$$ 3. **Rewrite the system:** $$\begin{cases} x - 3y - 3z = 7 \\ 3x + y = 7 \\ x + 2y - 2z = -6 \end{cases}$$ 4. **Express $y$ from the second equation:** $$3x + y = 7 \implies y = 7 - 3x$$ 5. **Substitute $y$ into the first and third equations:** - First equation: $$x - 3(7 - 3x) - 3z = 7$$ $$x - 21 + 9x - 3z = 7$$ $$10x - 3z - 21 = 7$$ $$10x - 3z = 28$$ - Third equation: $$x + 2(7 - 3x) - 2z = -6$$ $$x + 14 - 6x - 2z = -6$$ $$-5x - 2z + 14 = -6$$ $$-5x - 2z = -20$$ 6. **Rewrite the two equations with $x$ and $z$:** $$\begin{cases} 10x - 3z = 28 \\ -5x - 2z = -20 \end{cases}$$ 7. **Multiply the second equation by 2 to align $x$ coefficients:** $$-10x - 4z = -40$$ 8. **Add the first and modified second equations:** $$10x - 3z + (-10x - 4z) = 28 + (-40)$$ $$-7z = -12$$ 9. **Solve for $z$:** $$z = \frac{-12}{-7} = \frac{12}{7}$$ 10. **Substitute $z$ back into the first equation:** $$10x - 3 \times \frac{12}{7} = 28$$ $$10x - \frac{36}{7} = 28$$ $$10x = 28 + \frac{36}{7} = \frac{196}{7} + \frac{36}{7} = \frac{232}{7}$$ $$x = \frac{232}{7 \times 10} = \frac{232}{70} = \frac{116}{35}$$ 11. **Find $y$ using $y = 7 - 3x$:** $$y = 7 - 3 \times \frac{116}{35} = 7 - \frac{348}{35} = \frac{245}{35} - \frac{348}{35} = -\frac{103}{35}$$ **Final answer:** $$\boxed{\left(x, y, z\right) = \left(\frac{116}{35}, -\frac{103}{35}, \frac{12}{7}\right)}$$