Compute $|A|$ by expansion along row $1$:

$|A|=1(0-3)-1(-10-1)+2(-6-0)=-3+11-12=-4.$

Use the identity $(\mathrm{adj}A{)}^{-1}=\frac{A}{|A|}$. Let $B=2(\mathrm{adj}A{)}^{-1}=\frac{2A}{|A|}=\frac{2A}{-4}=-\frac{A}{2}$.

For a $3\times 3$ matrix, $\mathrm{adj}(\mathrm{adj}B)=|B{|}^{n-2}B=|B|B$.

$|B|={\left(-\frac{1}{2}\right)}^3|A|=-\frac{1}{8}\cdot (-4)=\frac{1}{2}.$

So $\mathrm{adj}(\mathrm{adj}B)=\frac{1}{2}\cdot \left(-\frac{A}{2}\right)=-\frac{A}{4}$.

Sum of entries of $A$: $1+1+2-2+0+1+1+3+5=12$. Therefore the required sum is $-\frac{12}{4}=-3$.