Let $$A$$ be a non-singular matrix of order 3. If $$\det(3 \text{ adj}(2 \text{ adj}((\det A)A))) = 3^{-13} \cdot 2^{-10}$$ and $$\det(3 \text{ adj}(2A)) = 2^m \cdot 3^n$$, then $$|3m + 2n|$$ is equal to ________

For an $$n \times n$$ matrix, $$\det(kA) = k^n \det(A)$$ and $$\det(\text{adj}(A)) = (\det A)^{n-1}$$. Here, $$n=3$$.

The expression is $$\det(3 \text{ adj}(2 \text{ adj}(d A)))$$.

$$\det(\text{adj}(dA)) = (d \cdot d^3)^2 = d^8$$.

$$\det(2 \text{ adj}(dA)) = 2^3 \cdot d^8$$.

$$\det(\text{adj}(2 \text{ adj}(dA))) = (2^3 d^8)^2 = 2^6 d^{16}$$.

$$\det(3 \text{ adj}(\dots)) = 3^3 \cdot 2^6 \cdot d^{16}$$.

Set $$3^3 \cdot 2^6 \cdot d^{16} = 3^{-13} \cdot 2^{-10} \implies d^{16} = 3^{-16} \cdot 2^{-16} \implies d = \frac{1}{6}$$.

$$\det(3 \text{ adj}(2A)) = 3^3 \cdot \det(\text{adj}(2A)) = 3^3 \cdot (\det(2A))^2 = 3^3 \cdot (2^3 d)^2 = 3^3 \cdot 2^6 \cdot d^2$$.

Substitute $$d = 2^{-1} \cdot 3^{-1}$$:

$$3^3 \cdot 2^6 \cdot (2^{-2} \cdot 3^{-2}) = 2^4 \cdot 3^1$$.

So, $$m = 4, n = 1$$.

$$|3(4) + 2(1)| = |12 + 2| = \mathbf{14}$$