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

$$\begin{gathered} \begin{array}{l}\left|\mathrm{A}\right|=1(6-1)-2(2\mathrm{\alpha }-1)+3(\mathrm{\alpha }-3)=5-4\mathrm{\alpha }+2+3\mathrm{\alpha }-9=2\\ \Rightarrow -\mathrm{\alpha }-2=2\Rightarrow \mathrm{\alpha }=-4\\ \left|2\mathrm{adj}\right(2\mathrm{adj}\left(2\mathrm{A}\right)\left)\right|=2^3|\mathrm{adj}(2\mathrm{adj}\left(2\mathrm{A}\right)\left)\right|\end{array} \\ \begin{array}{l}=2^3|2\mathrm{adj}(2\mathrm{A}){|}^2\\ ={\left.2^3{\left(2^3\right)}^2\left|\right|\mathrm{adj}\left(2\mathrm{A}\right)\right|}^2\\ =2^9|2\mathrm{A}{|}^4=2^9{\left(2^3\right)}^4|\mathrm{A}{|}^4\\ =2^9{2}^{12}{2}^4=2^{25}\\ ={\left(2^5\right)}^5={32}^5, \\ \mathrm{n}=5\\ 3\mathrm{n}+\mathrm{\alpha }=15-4=11\end{array} \end{gathered}$$