1. Use determinant formula for adjoint

For an $m \times m$ matrix $M$,

$$|\operatorname{adj}(M)| = |M|^{m-1}.$$

Here $m=3$, so

$$|\operatorname{adj}(M)| = |M|^2.$$

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2. Apply repeatedly to the given condition

Let $|A|=d$.

Then

$$|\operatorname{adj}(A)|=d^2.$$

Again,

$$|\operatorname{adj}(\operatorname{adj}(A))|=(d^2)^2=d^4.$$

And,

$$|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(A)))|=(d^4)^2=d^8.$$

Given

$$|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(A)))|=81,$$

so

$$d^8=81=3^4.$$

Hence

$$d=\pm 3^{1/2}=\pm \sqrt{3}.$$

But since exponents later are integers and determinant powers are used, we only need powers of $d$; using $d^8=81$ is sufficient.

Also,

$$|\operatorname{adj}(\operatorname{adj}(A))|=d^4.$$

Since $d^8=81$, we get

$$d^4=9.$$

Thus

$$|\operatorname{adj}(\operatorname{adj}(A))|=9.$$

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3. Use the condition defining $S$

We are given

$$\left(|\operatorname{adj}(\operatorname{adj}(A))|\right)^{\frac{(n-1)^2}{2}}=|A|^{3n^2-5n-4}.$$

Substitute $|\operatorname{adj}(\operatorname{adj}(A))|=d^4$:

$$(d^4)^{\frac{(n-1)^2}{2}}=d^{3n^2-5n-4}.$$

So

$$d^{2(n-1)^2}=d^{3n^2-5n-4}.$$

Thus

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

Now solve:

$$2(n^2-2n+1)=3n^2-5n-4$$ $$2n^2-4n+2=3n^2-5n-4$$ $$0=n^2-n-6$$ $$(n-3)(n+2)=0.$$

Therefore

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

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4. Compute $\sum_{n\in S}|A^{(n^2+n)}|$

For a $3\times 3$ matrix,

$$|A^k|=|A|^k=d^k.$$

But we need numerical value. Since both exponents obtained are even, we can use $d^4=9$.

For $n=3$:

$$n^2+n=9+3=12.$$

So

$$|A^{12}|=d^{12}=(d^4)^3=9^3=729.$$

For $n=-2$:

$$n^2+n=4-2=2.$$

So

$$|A^2|=d^2.$$

Since $d^4=9$, we get

$$d^2=3.$$

Hence

$$|A^2|=3.$$

Therefore,

$$\sum_{n\in S}|A^{(n^2+n)}|=729+3=732.$$

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5. Compare with stored answer

Derived answer: $732$.

Stored correct answer: D, i.e. $732$.

So they agree.