To find the expression $|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))|$, we break it down as follows:

Recognize that:

$ |2 \operatorname{adj}(3 A \operatorname{adj}(2 A))| = 2^3 |3A (\operatorname{adj}(2A))|^2 $

Apply properties of determinants:

$ = 2^3 (3^3)^2 |A|^2 \left|\operatorname{adj}(2A)\right|^2 $

Further simplify using $|\operatorname{adj}(B)| = |B|^{n-1}$ for a $3 \times 3$ matrix:

$ = 2^3 \cdot 3^6 \cdot 5^2 \cdot (|2A|^2)^2 $

Simplify $|2A|$:

$ = 2^3 \cdot 3^6 \cdot 5^2 \cdot (2^3)^4 \cdot |A|^4 $

Continue to simplify:

$ = 2^3 \cdot 3^6 \cdot 5^2 \cdot (2^3)^4 \cdot 5^4 $

Expand and combine powers:

$ = 2^{15} \cdot 3^6 \cdot 5^6 $

Therefore, $\alpha = 15$, $\beta = 6$, and $\gamma = 6$. So, $\alpha + \beta + \gamma = 15 + 6 + 6 = 27$.