Let $$A$$ and $$B$$ be two square matrices of order 3 such that $$|A| = 3$$ and $$|B| = 2$$. Then $$|A^T A(\text{adj}(2A))^{-1}(\text{adj}(4B))(\text{adj}(AB))^{-1}AA^T|$$ is equal to :

Given $$|A| = 3$$ and $$|B| = 2$$ for 3×3 matrices, we want to find $$|A^T A(\text{adj}(2A))^{-1}(\text{adj}(4B))(\text{adj}(AB))^{-1}AA^T|$$.

Some key properties for $$n \times n$$ matrices (with $$n = 3$$) are: $$|A^T| = |A|$$, $$|\text{adj}(M)| = |M|^{n-1} = |M|^2$$, $$|kM| = k^n|M| = k^3|M|$$, and $$|\text{adj}(kM)| = |kM|^2 = k^6|M|^2$$.

From these properties, one has $$|A^T| = 3$$ and $$|A| = 3$$. Moreover, $$|(\text{adj}(2A))^{-1}| = \frac{1}{|\text{adj}(2A)|} = \frac{1}{|2A|^2} = \frac{1}{(8 \times 3)^2} = \frac{1}{576}$$, while $$|\text{adj}(4B)| = |4B|^2 = (64 \times 2)^2 = 128^2 = 16384$$, and $$|(\text{adj}(AB))^{-1}| = \frac{1}{|AB|^2} = \frac{1}{(3 \times 2)^2} = \frac{1}{36}$$. Collecting these determinants gives

$$ = |A^T| \cdot |A| \cdot |(\text{adj}(2A))^{-1}| \cdot |\text{adj}(4B)| \cdot |(\text{adj}(AB))^{-1}| \cdot |A| \cdot |A^T| $$ $$ = 3 \times 3 \times \frac{1}{576} \times 16384 \times \frac{1}{36} \times 3 \times 3 $$ $$ = 81 \times \frac{16384}{576 \times 36} = 81 \times \frac{16384}{20736} $$ $$ = \frac{81 \times 16384}{20736} = \frac{1327104}{20736} = 64 $$

One can verify that $$81/20736 = 81/(81 \times 256) = 1/256$$, so $$16384/256 = 64$$. The correct answer is Option (4): 64.