The determinant of the adjoint of a matrix \(A\) is related to the determinant of the matrix as follows:
\[
\text{det(adj}(A)) = (\text{det}(A))^{n-1}
\]
For a 3×3 matrix, this becomes:
\[
\text{det(adj}(A)) = (\text{det}(A))^2
\]
Also, \(\text{det}(kA) = k^n \cdot \text{det}(A)\) for an \(n \times n\) matrix, so:
\[
\text{det}(3 \, \text{adj}(2A)) = 3^3 \cdot 2^3 \cdot (\text{det}(A))^2 = 27 \cdot 8 \cdot 5^2 = 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma}
\]
Thus, \(\alpha = 3\), \(\beta = 3\), and \(\gamma = 4\).
So, \(\alpha + \beta + \gamma = 3 + 3 + 4 = 27\).