If $$A$$ is a square matrix of order 3 such that $$\det(A) = 3$$ and $$\det(\text{adj}(-4 \text{adj}(-3 \text{adj}(3 \text{adj}((2A)^{-1}))))) = 2^m 3^n$$, then $$m + 2n$$ is equal to :

$$|\text{adj}(kM)| = k^{n(n-1)} |\text{adj}(M)| = k^{6}|M|^2$$ for order $$3$$, and $$|\text{adj}(M)| = |M|^2$$.

Inner term: $$|(2A)^{-1}| = \frac{1}{8|A|} = \frac{1}{24}$$.

$$|3\text{adj}((2A)^{-1})| = 3^3 \cdot \left(\frac{1}{24}\right)^2 = \frac{27}{576} = \frac{3}{64}$$.

$$|-3\text{adj}(\dots)| = (-3)^3 \cdot \left(\frac{3}{64}\right)^2 = -\frac{243}{4096}$$.

$$|-4\text{adj}(\dots)| = (-4)^3 \cdot \left(-\frac{243}{4096}\right)^2 = -64 \cdot \frac{59049}{16777216} = -\frac{59049}{262144}$$.

Final determinant is the square of the previous result: $$\left(-\frac{59049}{262144}\right)^2 = \frac{3^{20}}{2^{36}} = 2^{-36} \cdot 3^{20}$$.

Matching parameters: $$m = -36$$, $$n = 20 \implies m + 2n = -36 + 40 = \mathbf{4}$$