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

$$ \begin{aligned} & A=\left[\begin{array}{lll} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{array}\right] \\\\ & |A|=2 \end{aligned} $$ <br/><br/>$$ \begin{aligned} \Rightarrow&1(6-1)-2(2 \alpha-1)+3(\alpha-3)=2\\\\ \Rightarrow&5-4 \alpha+2+3 \alpha-9=2\\\\ \Rightarrow&-\alpha-4=0\\\\ \Rightarrow&\alpha=-4 \end{aligned} $$ <br/><br/>Now, $\mathrm{|2\,adj\,(2\,adj\,(2A))| = {32^n}}$ <br/><br/>$\Rightarrow$ $2^3|\operatorname{adj}(2 \operatorname{adj}(2 A))|$ = ${32^n}$ <br/><br/>$$ \begin{aligned} \Rightarrow & 8|\operatorname{Adj}(2 \operatorname{Adj}(2 \mathrm{~A}))| = {32^n}\\\\ \Rightarrow & 8\left|\operatorname{Adj}\left(2 \times 2^2 \operatorname{Adj}(\mathrm{A})\right)\right| = {32^n} \\\\ \Rightarrow & 8\left|\operatorname{Adj}\left(2^3 \operatorname{AdjA}\right)\right| = {32^n} \\\\ \Rightarrow & 8\left|2^6 \operatorname{Adj}(\operatorname{AdjA})\right| = {32^n} \\\\ \Rightarrow & 2^3\left(2^6\right)^3|\operatorname{Adj}(\operatorname{Adj})| = {32^n} \\\\ \Rightarrow & 2^3 \cdot 2^{18}|\mathrm{~A}|^4 = {32^n} \\\\ \Rightarrow& 2^{21} \cdot 2^4 = {32^n}\\\\ \Rightarrow& 2^{25} = {32^n}\\\\ \Rightarrow& \left(2^5\right)^5 = {32^n}\\\\ \Rightarrow& (32)^5 = {32^n} \end{aligned} $$ <br/><br/>$$ \begin{array}{ll} \therefore n=5 \\\\ \Rightarrow 3 n+\alpha=15-4=11 \end{array} $$