1. **State the problem:** We need to find the value of $a$ for which the system of simultaneous equations has a unique solution: $$\begin{cases} x + y + z = 2 \\ x + 2y - 3z = 5 \\ 2x + ay - az = 3 \end{cases}$$ 2. **Recall the condition for a unique solution:** A system of linear equations has a unique solution if and only if the coefficient matrix is invertible, i.e., its determinant is non-zero. 3. **Write the coefficient matrix:** $$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & a & -a \end{bmatrix}$$ 4. **Calculate the determinant of $A$:** $$\det(A) = 1 \cdot \begin{vmatrix} 2 & -3 \\ a & -a \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & -3 \\ 2 & -a \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 2 & a \end{vmatrix}$$ Calculate each minor: $$\begin{aligned} M_1 &= (2)(-a) - (a)(-3) = -2a + 3a = a \\ M_2 &= (1)(-a) - (2)(-3) = -a + 6 = 6 - a \\ M_3 &= (1)(a) - (2)(2) = a - 4 \end{aligned}$$ So, $$\det(A) = 1 \cdot a - 1 \cdot (6 - a) + 1 \cdot (a - 4) = a - (6 - a) + (a - 4)$$ Simplify: $$a - 6 + a + a - 4 = 3a - 10$$ 5. **Set determinant not equal to zero for unique solution:** $$3a - 10 \neq 0$$ Solve for $a$: $$3a \neq 10 \implies a \neq \frac{10}{3}$$ 6. **Conclusion:** The system has a unique solution for all values of $a$ except $a = \frac{10}{3}$. **Final answer:** $$a \neq \frac{10}{3}$$