Determinant of Adjoint Matrix given Matrix Determinant Condition

To find $\text{det}(\text{adj}(A))$, we first determine the adjugate matrix $\text{adj}(A)$ of $A$.

Given the matrix: $A = \begin{bmatrix} \alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix}$

The matrix of cofactors of $A$, denoted by $C$, is calculated as: $C_{11} = 15, \quad C_{12} = -10, \quad C_{13} = 8$ $C_{21} = 3, \quad C_{22} = 5\alpha, \quad C_{23} = -4\alpha$ $C_{31} = -6, \quad C_{32} = 4, \quad C_{33} = 3\alpha - 2$

Thus, the adjugate of $A$ is the transpose of the cofactor matrix: $\text{adj}(A) = C^T = \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix}$

Now, matrix $B$ is given by: $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix} = \begin{bmatrix} 16 & 3 & -6 \\ -10 & 0 & 4 \\ 8 & 0 & \alpha - 2 \end{bmatrix}$

Next, we evaluate $\text{det}(B)$ by expanding along the second column: $\text{det}(B) = -3 \begin{vmatrix} -10 & 4 \\ 8 & \alpha - 2 \end{vmatrix} = -3 \left[ -10(\alpha - 2) - 32 \right] = -3 (-10\alpha - 12) = 30\alpha + 36$

Given $\text{det}(B) = 66$: $30\alpha + 36 = 66 \implies 30\alpha = 30 \implies \alpha = 1$

Substitute $\alpha = 1$ into matrix $A$ to compute its determinant: $\text{det}(A) = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{vmatrix} = 1(15 - 0) - 1(10 - 0) + 2(8 - 0) = 15 - 10 + 16 = 21$

For a $3 \times 3$ matrix $A$, the determinant of its adjugate is given by: $\text{det}(\text{adj}(A)) = (\text{det}(A))^{3-1} = (\text{det}(A))^2$

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

Therefore, the correct option is C.