Let A be a $$3 \times 3$$ matrix such that $$|\text{adj}(\text{adj}(\text{adj } A))| = 81$$. If $$S = \{n \in \mathbb{Z} : (|\text{adj}(\text{adj } A)|)^{\frac{(n-1)^2}{2}} = |A|^{(3n^2 - 5n - 4)}\}$$, then $$\sum_{n \in S} |A^{(n^2+n)}|$$ is equal to

Given that $$A$$ is a $$3 \times 3$$ matrix. For an $$n \times n$$ matrix, $$|\text{adj}(A)| = |A|^{n-1}$$.

For $$n = 3$$: $$|\text{adj}(A)| = |A|^2$$, $$|\text{adj}(\text{adj}(A))| = |A|^{(n-1)^2} = |A|^4$$, and $$|\text{adj}(\text{adj}(\text{adj}(A)))| = |A|^{(n-1)^3} = |A|^8$$.

Given $$|A|^8 = 81 = 3^4$$, so $$|A|^2 = 81^{1/4} = 3$$.

Now for the set $$S$$, we need: $$|\text{adj}(\text{adj}(A))|^{\frac{(n-1)^2}{2}} = |A|^{3n^2 - 5n - 4}$$

$$|A|^{4 \cdot \frac{(n-1)^2}{2}} = |A|^{3n^2 - 5n - 4}$$

$$|A|^{2(n-1)^2} = |A|^{3n^2 - 5n - 4}$$

Since $$|A| \neq 0$$ and $$|A| \neq \pm 1$$, equating exponents:

$$2(n^2 - 2n + 1) = 3n^2 - 5n - 4$$

$$2n^2 - 4n + 2 = 3n^2 - 5n - 4$$

$$n^2 - n - 6 = 0$$, giving $$(n-3)(n+2) = 0$$

So $$S = \{3, -2\}$$.

$$\sum_{n \in S} |A^{n^2+n}| = |A|^{9+3} + |A|^{4-2} = |A|^{12} + |A|^{2}$$

Since $$|A|^2 = 3$$: $$|A|^{12} = (|A|^2)^6 = 3^6 = 729$$ and $$|A|^2 = 3$$.

Sum $$= 729 + 3 = 732$$.

Hence, the correct answer is Option D.