Given that $A=\left[\begin{array}{lll}1& 2& 3\\ \alpha & 3& 1\\ 1& 1& 2\end{array}\right]$

$\Rightarrow \left|A\right|=2$

$\Rightarrow 1\left(6-1\right)-\alpha -7=2$

$\Rightarrow -\alpha -2=2$

$\Rightarrow \alpha =-4$

$\Rightarrow |2adj(2adj\left(2A\right)\left)\right|={32}^n$

We know that $\left|kA\right|=k^n\left|A\right|$, $adj(adj(A\left)\right)={\left|A\right|}^{{\left(n-1\right)}^2}$ and $\mathrm{adj}\left(kA\right)\mathit{=}{k}^{n\mathit{-}\mathit{1}}\mathrm{adj}\left(A\right)$.

$=2^3|adj(2adj\left(2A\right)\left)\right|$

$=2^3|2^{3-1}adj(adj\left(2A\right)\left)\right|$

$=2^3·\left(4{)}^3\right|adj(adj(2A\left)\right)|$

$=2^9|2A{|}^{(2{)}^2}$

$=2^9·|2A{|}^4$

$=2^9·2^{12}|A{|}^4$

$=2^{21}·2^4=2^{25}=(32{)}^n=(2{)}^{5n}$

$\therefore n=5$

$\Rightarrow 3n+\alpha =15-4=11$

Hence this is the correct option.