If $$\mathrm{A}=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{ccc}5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{array}\right]$$, then $|\operatorname{adj}(\operatorname{adj}(2 \mathrm{~A}))|$ is equal to :

Given that <br/><br/>$$ \begin{aligned} & A=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{lll} 5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 ! \end{array}\right] \\\\ & \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !}\left|\begin{array}{lll} 5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 ! \end{array}\right| \\\\ & \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !} \times 5 ! 6 ! 7 !\left|\begin{array}{lll} 1 & 6 & 7 \times 6 \\ 1 & 7 & 8 \times 7 \\ 1 & 8 & 9 \times 8 \end{array}\right| \end{aligned} $$ <br/><br/>$$ \begin{aligned} & R_3 \rightarrow R_3-R_2 \text { and } R_2 \rightarrow R_2-R_1 \\\\ & |A|=\left|\begin{array}{lll} 1 & 6 & 42 \\ 0 & 1 & 14 \\ 0 & 1 & 16 \end{array}\right|=2 \\\\ & |\operatorname{adj}(\operatorname{adj}(2 A))|=|2 A|^{(n-1)^2} \\\\ & =|2 A|^4=\left(2^3|A|\right)^4=2^{12}|A|^4=2^{16} \end{aligned} $$