Let $$A = \begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}$$. If $$\det(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n$$, $$m, n \in \mathbb{N}$$, then $$m + n$$ is equal to

We must evaluate $$\det\bigl(\text{adj}\bigl(\text{adj}(3A)\bigr)\bigr)$$, where the given matrix is
$$A=\begin{bmatrix}2 & 2+p & 2+p+q\\ 4 & 6+2p & 8+3p+2q\\ 6 & 12+3p & 20+6p+3q\end{bmatrix}.$$

Case 1 : Relating adjugates to determinants

For any non-singular $$n\times n$$ matrix $$M$$, two standard identities are

$$\det\bigl(\text{adj}(M)\bigr)=(\det M)^{\,n-1}$$

$$\text{adj}\bigl(\text{adj}(M)\bigr)=(\det M)^{\,n-2}\,M$$  for $$n\gt1$$.

Here $$n=3$$ and $$M=3A$$. Therefore

$$\text{adj}\bigl(\text{adj}(3A)\bigr)=\bigl(\det(3A)\bigr)^{\,3-2}\,(3A)=\det(3A)\,(3A).$$

Case 2 : Determinant of the obtained matrix

Let $$B=\det(3A)\,(3A)$$. Since multiplying a $$3\times3$$ matrix by a scalar $$c$$ multiplies its determinant by $$c^{\,3}$$, we get

$$\det(B)=\bigl(\det(3A)\bigr)^{3}\,\det(3A)=\bigl(\det(3A)\bigr)^{4}.$$

Case 3 : Evaluating $$\det(3A)$$

First compute $$\det(A)$$. Apply the row operations $$R_2\leftarrow R_2-2R_1$$ and $$R_3\leftarrow R_3-3R_1$$:

$$\begin{bmatrix} 2 & 2+p & 2+p+q\\ 4 & 6+2p & 8+3p+2q\\ 6 & 12+3p & 20+6p+3q \end{bmatrix} \;\longrightarrow\; \begin{bmatrix} 2 & 2+p & 2+p+q\\ 0 & 2 & 4+p\\ 0 & 6 & 14+3p \end{bmatrix}.$$ Only elementary row replacements were used, so the determinant is unchanged. Expanding along the first column,

$$\det(A)=2\begin{vmatrix}2 & 4+p\\ 6 & 14+3p\end{vmatrix} =2\bigl[(2)(14+3p)-(4+p)(6)\bigr]$$

$$=2\bigl[(28+6p)-(24+6p)\bigr]=2\,(4)=8.$$

Hence $$\det(A)=8=2^{3}.$$

Now, $$\det(3A)=3^{3}\det(A)=27\times8=216=2^{3}\,3^{3}.$$

Case 4 : Final value of the required determinant

Substituting $$\det(3A)=2^{3}3^{3}$$ into Case 2 gives

$$\det\bigl(\text{adj}(\text{adj}(3A))\bigr)=\bigl(2^{3}3^{3}\bigr)^{4}=2^{12}\,3^{12}.$$

Thus $$m=12,\;n=12$$ and

$$m+n=24.$$

The correct choice is Option B (24).