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

<p>$|A|=\left|\begin{array}{ccc}2 & 2+p & 2+p+q \\ 4 & 6+2 p & 8+3 p+2 q \\ 6 & 12+3 p & 20+6 p+3 q\end{array}\right|$</p> <p>$$\begin{aligned} &\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_2-\mathrm{C}_1 \times \frac{\mathrm{q}}{2}\\ &\text { Then } \mathrm{C}_3 \rightarrow \mathrm{C}_2-\mathrm{C}_1 \mathrm{X}\left(1+\frac{\mathrm{p}}{2}\right)\\ &\Rightarrow|\mathrm{A}|=\left|\begin{array}{ccc} 2 & 0 & 0 \\ 4 & 2 & 2+\mathrm{p} \\ 6 & 6 & 8+3 \mathrm{p} \end{array}\right| \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow|\mathrm{A}|=2(16+6 \mathrm{p}-12-6 \mathrm{p})=8=2^3 \\ & |\operatorname{adj}(\operatorname{adj}(3 \mathrm{~A}))|=|3 \mathrm{~A}|^{(3-1)^2}=|3 \mathrm{~A}|^4 \\ & =\left(3^3|\mathrm{~A}|\right)^4=\left(3^3 \times 2^3\right)^4=2^{12} \times 3^{12} \\ & \Rightarrow \mathrm{~m}+\mathrm{n}=24 \end{aligned}$$</p>