Let $B=2\,\operatorname{adj}(3A\operatorname{adj}(2A)).$ We need to find $|B|=\left|2\,\operatorname{adj}(3A\operatorname{adj}(2A))\right|.$

First simplify the inner determinant.

For a $3\times 3$ matrix:

• $|kA|=k^3|A|$
• $|\operatorname{adj}(M)|=|M|^{n-1}$ for an $n\times n$ matrix, so for $n=3$, $|\operatorname{adj}(M)|=|M|^2.$

Since $|A|=5$, $|2A|=2^3|A|=8\cdot 5=40.$

Hence, $|\operatorname{adj}(2A)|=|2A|^2=40^2=1600.$

Also, $|3A|=3^3|A|=27\cdot 5=135.$

Now compute $|3A\operatorname{adj}(2A)|=|3A|\cdot|\operatorname{adj}(2A)|=135\cdot 1600.$

So, $|3A\operatorname{adj}(2A)|=216000.$

Prime factorize: $216000=216\cdot 1000=2^4\cdot 3^3\cdot 5^3.$

Now use the adjoint determinant formula again.

Let $M=3A\operatorname{adj}(2A).$ Then $|\operatorname{adj}(M)|=|M|^2=(2^4\cdot 3^3\cdot 5^3)^2=2^8\cdot 3^6\cdot 5^6.$

Now multiply by the outer scalar $2$.

Since $\operatorname{adj}(M)$ is also a $3\times 3$ matrix, $|2\operatorname{adj}(M)|=2^3\,|\operatorname{adj}(M)|.$

Therefore, $|2\operatorname{adj}(M)|=2^3\cdot 2^8\cdot 3^6\cdot 5^6=2^{11}\cdot 3^6\cdot 5^6.$

Compare with $2^\alpha\cdot 3^\beta\cdot 5^\gamma.$ We get $\alpha=11,\quad \beta=6,\quad \gamma=6.$

Hence, $\alpha+\beta+\gamma=11+6+6=23.$

So the derived answer is $\boxed{23}.$

Comparison with stored answer:

• Stored correct answer: B = 27
• Derived answer: 23

Therefore, I do not agree with the stored answer.

Quick verification by an alternative compact method:

$\operatorname{adj}(2A)=2^{3-1}\operatorname{adj}(A)=4\operatorname{adj}(A).$ So, $3A\operatorname{adj}(2A)=3A\cdot 4\operatorname{adj}(A)=12A\operatorname{adj}(A)=12|A|I=60I.$ Then, $\operatorname{adj}(60I)=60^2 I$ (for a $3\times 3$ matrix), hence $2\operatorname{adj}(60I)=2\cdot 60^2 I=7200I.$ Therefore, $|2\operatorname{adj}(60I)|=(7200)^3=(2^5\cdot 3^2\cdot 5^2)^3=2^{15}3^65^6.$

This reveals an important correction: the previous step using only determinant properties missed the fact that the matrix itself simplifies to $60I$, and scaling occurs before determinant evaluation.

Thus actually, $\alpha=15,\ \beta=6,\ \gamma=6,$ so $\alpha+\beta+\gamma=27.$

Final correct answer: $\boxed{27}$ which matches option B.