Let A = bmatrix α & 1 & 2 2 & 3 & 0 0 & 4 & 5 bmatrix and B =...

The cofactor matrix of $A$ is calculated as follows:

$C_{11} = 15$, $C_{12} = -10$, $C_{13} = 8$

$C_{21} = 3$, $C_{22} = 5\alpha$, $C_{23} = -4\alpha$

$C_{31} = -6$, $C_{32} = 4$, $C_{33} = 3\alpha - 2$

The adjoint of $A$ is the transpose of the cofactor matrix:

$\text{adj}(A) = \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix}$

The matrix $B$ is given by:

$B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix} = \begin{bmatrix} 16 & 3 & -6 \\ -10 & 0 & 4 \\ 8 & 0 & \alpha - 2 \end{bmatrix}$

Expanding the determinant of $B$ along the second column:

$\det(B) = -3 \begin{vmatrix} -10 & 4 \\ 8 & \alpha - 2 \end{vmatrix} = -3(-10(\alpha - 2) - 32) = -3(-10\alpha - 12) = 30\alpha + 36$

Given $\det(B) = 66$, we get:

$30\alpha + 36 = 66 \Rightarrow 30\alpha = 30 \Rightarrow \alpha = 1$

The determinant of $A$ is:

$\det(A) = \alpha(15 - 0) - 1(10 - 0) + 2(8 - 0) = 15\alpha + 6$

Substituting $\alpha = 1$:

$\det(A) = 15(1) + 6 = 21$

Using the property $\det(\text{adj}(A)) = (\det(A))^{n-1}$ for a $3 \times 3$ matrix:

$\det(\text{adj}(A)) = (\det(A))^2 = (21)^2 = 441$

Answer: $441$