Let $$\alpha\beta \neq 0$$ and $$A = \begin{bmatrix} \beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2\alpha \end{bmatrix}$$. If $$B = \begin{bmatrix} 3\alpha & -9 & 3\alpha \\ -\alpha & 7 & -2\alpha \\ -2\alpha & 5 & -2\beta \end{bmatrix}$$ is the matrix of cofactors of the elements of $$A$$, then $$\det(AB)$$ is equal to :

Matrix $$B$$ is stated to be the matrix of cofactors of the elements of $$A$$.
For any square matrix, if $$C$$ is its cofactor matrix, then

$$A\,C = \det(A)\,I_3 \qquad -(1)$$

because the scalar product of one row of $$A$$ with the cofactors of any different row vanishes, while the product with the cofactors of the same row equals $$\det(A)$$.

Taking determinant on both sides of $$(1)$$ gives

$$\det(A\,C)=\det(A)^3 \qquad -(2)$$

Since $$B$$ is that cofactor matrix, $$\det(AB)=\det(A)^3$$.
Thus our task reduces to finding $$\det(A)$$. To do this we first determine $$\alpha$$ and $$\beta$$ from the fact that the cofactors of $$A$$ equal the corresponding entries of $$B$$.

The needed cofactors of the first row of $$A=\begin{bmatrix}\beta & \alpha & 3\\ \alpha & \alpha & \beta\\ -\beta & \alpha & 2\alpha\end{bmatrix}$$ are computed below.

Case 1: Cofactor $$C_{11}$$ (remove row 1, column 1)

$$\begin{vmatrix}\alpha & \beta\\ \alpha & 2\alpha\end{vmatrix} = 2\alpha^2-\alpha\beta \quad\Longrightarrow\quad C_{11}=2\alpha^2-\alpha\beta$$

Given $$B_{11}=3\alpha$$, therefore

$$2\alpha^2-\alpha\beta = 3\alpha \quad\Longrightarrow\quad 2\alpha-\beta = 3 \qquad -(3)$$

Case 2: Cofactor $$C_{12}$$ (remove row 1, column 2)

Minor $$M_{12}= \begin{vmatrix}\alpha & \beta\\ -\beta & 2\alpha\end{vmatrix} =2\alpha^2+\beta^2$$
Cofactor $$C_{12}=(-1)^{1+2}M_{12}=-(2\alpha^2+\beta^2)$$

Given $$B_{12}=-9$$, therefore

$$-(2\alpha^2+\beta^2)=-9 \quad\Longrightarrow\quad 2\alpha^2+\beta^2 = 9 \qquad -(4)$$

Case 3: Cofactor $$C_{13}$$ (remove row 1, column 3)

$$\begin{vmatrix}\alpha & \alpha\\ -\beta & \alpha\end{vmatrix} =\alpha^2+\alpha\beta \quad\Longrightarrow\quad C_{13}= \alpha^2+\alpha\beta$$

Given $$B_{13}=3\alpha$$, therefore

$$\alpha^2+\alpha\beta = 3\alpha \quad\Longrightarrow\quad \alpha+\beta = 3 \qquad -(5)$$

Solving $$(3)$$ and $$(5)$$ simultaneously:

From $$(5)$$, $$\beta = 3-\alpha$$. Substituting in $$(3)$$:

$$2\alpha-(3-\alpha)=3 \quad\Longrightarrow\quad 3\alpha=6 \quad\Longrightarrow\quad \alpha=2$$

Then $$\beta = 3-\alpha = 1$$. These values also satisfy $$(4)$$ since $$2(2)^2+1^2 = 8+1 = 9$$.

Determinant of $$A$$ with $$\alpha=2,\;\beta=1$$

$$A=\begin{bmatrix}1 & 2 & 3\\ 2 & 2 & 1\\ -1 & 2 & 4\end{bmatrix}$$

Expanding along the first row,

$$\det(A)= 1\bigl(2\cdot4-1\cdot2\bigr) -2\bigl(2\cdot4-1\cdot(-1)\bigr) +3\bigl(2\cdot2-2\cdot(-1)\bigr)$$

$$\det(A)=1(8-2)-2(8+1)+3(4+2)=6-18+18=6 \qquad -(6)$$

Determinant of $$AB$$

Using $$(2)$$ and $$(6)$$,

$$\det(AB)=\det(A)^3 = 6^3 = 216$$

Therefore the correct option is Option B (216).