Let the determinant of a square matrix A of order m be m - n ,...

We have been given that

$\left|A\right| = m - n$, where $4m + n = 22 . . . . . . . \left(i\right)$

and $17m + 4n = 93 . . . . . . . \left(ii\right)$

Solving \left(i\right)$and \left(ii\right)$

$m = 5, n = 2$

Order $= 5$

$⇒ \left|A\right| = 3$

We know that \left|adjA\right| = {\left|A\right|}^{w - 1}$\left|adj\left(adjA\right)\right| = {\left|A\right|}^{{\left(w - 1\right)}^{2}}$ where $w \times w$ is the order of the matrix.

$∴ det\left(nadj\left(adj\left(mA\right)\right)\right)$

$= \left|2adj\left(adj\left(5A\right)\right)\right|$

$= {2}^{5}{\left|5A\right|}^{16}$

$= {2}^{5}{5}^{80}{\left|A\right|}^{16} = {2}^{5} \cdot {3}^{16} \cdot {5}^{80}$

$= {3}^{11}{5}^{80}{6}^{5}$

So, $a + b + c = 96$

Hence this is the required option.