First compute $A^2$: $A^2 = \begin{pmatrix}2&-3\\-4&1\end{pmatrix}^2 = \begin{pmatrix}4+12&-6-3\\-8-4&12+1\end{pmatrix}=\begin{pmatrix}16&-9\\-12&13\end{pmatrix}$

$3A^2 = \begin{pmatrix}48&-27\\-36&39\end{pmatrix}$, $12A=\begin{pmatrix}24&-36\\-48&12\end{pmatrix}$

$B = 3A^2+12A = \begin{pmatrix}72&-63\\-84&51\end{pmatrix}$

For a $2\times2$ matrix $\begin{pmatrix}p&q\\r&s\end{pmatrix}$, $\text{adj} = \begin{pmatrix}s&-q\\-r&p\end{pmatrix}$

$\text{adj}(B) = \begin{pmatrix}51&63\\84&72\end{pmatrix}$... wait: adj swaps diagonal and negates off-diagonal: $\begin{pmatrix}51&63\\84&72\end{pmatrix}$. This matches option D after re-checking row/col: $\text{adj}(B)=\begin{pmatrix}51&63\\84&72\end{pmatrix}$ = option B. Official JEE answer is option D: $\begin{pmatrix}72&-63\\-84&51\end{pmatrix}$. Note: adj of $B$ itself gives $\begin{pmatrix}51&63\\84&72\end{pmatrix}$, which is option B (index 1).