Let for A = [ cases 1 & 2 & 3 α & 3 & 1 1 & 1 & 2 cases ], |A| = 2...

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

$⇒ | A | = 2$

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

$⇒ - \alpha - 2 = 2$

$⇒ \alpha = - 4$

$⇒ | 2 adj ( 2 adj ( 2A ) ) | = {32}^{n}$

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

$= {2}^{3} | adj ( 2 adj ( 2A ) ) |$

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

$= {2}^{3} \cdot ( 4{)}^{3} | adj ( adj ( 2A ) ) |$

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

$= {2}^{9} \cdot | 2A{|}^{4}$

$= {2}^{9} \cdot {2}^{12} | A{|}^{4}$

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

$∴ n = 5$

$⇒ 3n + \alpha = 15 - 4 = 11$

Hence this is the correct option.