1. **State the problem:** Solve the system of equations using Cramer's rule: $$\begin{cases} 2x - 3y + z = 17 \\ x + 3y - 2z = -13 \\ x - y - z = 0 \end{cases}$$ 2. **Write the coefficient matrix $A$ and constants vector:** $$A = \begin{bmatrix} 2 & -3 & 1 \\ 1 & 3 & -2 \\ 1 & -1 & -1 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 17 \\ -13 \\ 0 \end{bmatrix}$$ 3. **Calculate determinant of $A$, denoted $\Delta$:** $$\Delta = \begin{vmatrix} 2 & -3 & 1 \\ 1 & 3 & -2 \\ 1 & -1 & -1 \end{vmatrix}$$ Expanding along the first column: $$\Delta = 2 \begin{vmatrix} 3 & -2 \\ -1 & -1 \end{vmatrix} - 1 \begin{vmatrix} -3 & 1 \\ -1 & -1 \end{vmatrix} + 1 \begin{vmatrix} -3 & 1 \\ 3 & -2 \end{vmatrix}$$ Calculate each minor: $$\begin{aligned} M_1 &= (3)(-1) - (-2)(-1) = -3 - 2 = -5 \\ M_2 &= (-3)(-1) - (1)(-1) = 3 + 1 = 4 \\ M_3 &= (-3)(-2) - (1)(3) = 6 - 3 = 3 \end{aligned}$$ So, $$\Delta = 2(-5) - 1(4) + 1(3) = -10 - 4 + 3 = -11$$ 4. **Calculate $\Delta_x$ by replacing first column of $A$ with $\mathbf{b}$:** $$\Delta_x = \begin{vmatrix} 17 & -3 & 1 \\ -13 & 3 & -2 \\ 0 & -1 & -1 \end{vmatrix}$$ Expanding along first column: $$\Delta_x = 17 \begin{vmatrix} 3 & -2 \\ -1 & -1 \end{vmatrix} - (-13) \begin{vmatrix} -3 & 1 \\ -1 & -1 \end{vmatrix} + 0 \begin{vmatrix} -3 & 1 \\ 3 & -2 \end{vmatrix}$$ Calculate minors (already found above): $$\begin{aligned} M_1 &= -5 \\ M_2 &= 4 \end{aligned}$$ So, $$\Delta_x = 17(-5) + 13(4) + 0 = -85 + 52 = -33$$ 5. **Calculate $\Delta_y$ by replacing second column of $A$ with $\mathbf{b}$:** $$\Delta_y = \begin{vmatrix} 2 & 17 & 1 \\ 1 & -13 & -2 \\ 1 & 0 & -1 \end{vmatrix}$$ Expanding along first column: $$\Delta_y = 2 \begin{vmatrix} -13 & -2 \\ 0 & -1 \end{vmatrix} - 1 \begin{vmatrix} 17 & 1 \\ 0 & -1 \end{vmatrix} + 1 \begin{vmatrix} 17 & 1 \\ -13 & -2 \end{vmatrix}$$ Calculate minors: $$\begin{aligned} M_1 &= (-13)(-1) - (-2)(0) = 13 - 0 = 13 \\ M_2 &= 17(-1) - 1(0) = -17 - 0 = -17 \\ M_3 &= 17(-2) - 1(-13) = -34 + 13 = -21 \end{aligned}$$ So, $$\Delta_y = 2(13) - 1(-17) + 1(-21) = 26 + 17 - 21 = 22$$ 6. **Calculate $\Delta_z$ by replacing third column of $A$ with $\mathbf{b}$:** $$\Delta_z = \begin{vmatrix} 2 & -3 & 17 \\ 1 & 3 & -13 \\ 1 & -1 & 0 \end{vmatrix}$$ Expanding along first column: $$\Delta_z = 2 \begin{vmatrix} 3 & -13 \\ -1 & 0 \end{vmatrix} - 1 \begin{vmatrix} -3 & 17 \\ -1 & 0 \end{vmatrix} + 1 \begin{vmatrix} -3 & 17 \\ 3 & -13 \end{vmatrix}$$ Calculate minors: $$\begin{aligned} M_1 &= 3(0) - (-13)(-1) = 0 - 13 = -13 \\ M_2 &= (-3)(0) - 17(-1) = 0 + 17 = 17 \\ M_3 &= (-3)(-13) - 17(3) = 39 - 51 = -12 \end{aligned}$$ So, $$\Delta_z = 2(-13) - 1(17) + 1(-12) = -26 - 17 - 12 = -55$$ 7. **Find the solutions using Cramer's rule:** $$x = \frac{\Delta_x}{\Delta} = \frac{-33}{-11} = 3$$ $$y = \frac{\Delta_y}{\Delta} = \frac{22}{-11} = -2$$ $$z = \frac{\Delta_z}{\Delta} = \frac{-55}{-11} = 5$$ **Final answer:** $$\boxed{(x, y, z) = (3, -2, 5)}$$