1. The problem asks for the condition on $a$ such that the matrix $$ A = \begin{bmatrix} -6 & a & a \\ 2 & -1 & 0 \\ -2 & 1 & 5 \end{bmatrix} $$ is invertible. 2. A matrix is invertible if and only if its determinant is nonzero. 3. Calculate the determinant of $A$: $$\det(A) = -6 \cdot \det\begin{bmatrix} -1 & 0 \\ 1 & 5 \end{bmatrix} - a \cdot \det\begin{bmatrix} 2 & 0 \\ -2 & 5 \end{bmatrix} + a \cdot \det\begin{bmatrix} 2 & -1 \\ -2 & 1 \end{bmatrix}$$ 4. Compute each minor determinant: $$\det\begin{bmatrix} -1 & 0 \\ 1 & 5 \end{bmatrix} = (-1)(5) - (0)(1) = -5$$ $$\det\begin{bmatrix} 2 & 0 \\ -2 & 5 \end{bmatrix} = (2)(5) - (0)(-2) = 10$$ $$\det\begin{bmatrix} 2 & -1 \\ -2 & 1 \end{bmatrix} = (2)(1) - (-1)(-2) = 2 - 2 = 0$$ 5. Substitute back: $$\det(A) = -6(-5) - a(10) + a(0) = 30 - 10a$$ 6. For $A$ to be invertible, $\det(A) \neq 0$: $$30 - 10a \neq 0$$ 7. Solve for $a$: $$30 \neq 10a$$ $$a \neq 3$$ 8. Therefore, the matrix $A$ is invertible if and only if $a \neq 3$. Final answer: (a) $a \neq 3$