Let for A = [1, 2, 3; α, 3, 1; 1, 1, 2],| A |=2. If |2 adj (2 adj (2 A ))|=32^n, then 3 n+α is equal to

Let for $\displaystyle \mathrm{A}=\left[\begin{array}{lll}1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2\end{array}\right],|\mathrm{A}|=2$. If $\displaystyle |2 \operatorname{adj}(2 \operatorname{adj}(2 \mathrm{~A}))|=32^n$, then $\displaystyle 3 n+\alpha$ is equal to