Let M denote the set of all real matrices of order $$3\times 3$$ and let$$S=\left\{-3,-2,-1,1,2\right\}$$. Let
$$S_{1}=\left\{A=[a_{ij}] \in M : A=A^{T}\text{ and }a_{ij} \in S,\forall i,j\right\},$$
$$S_{2}=\left\{A=[a_{ij}] \in M : A=-A^{T}\text{ and }a_{ij} \in S,\forall i,j\right\},$$
$$S_{3}=\left\{A=[a_{ij}] \in M : a_{11}+a_{22}+a_{33}=0\text{ and }a_{ij} \in S,\forall i,j\right\},$$
If $$n(S_{1}\cup_{2} US_{3})=125\alpha$$, then $$alpha$$ equals___________

We need to find $$\alpha$$ where $$n(S_1 \cup S_2 \cup S_3) = 125\alpha$$. Since $$S = \{-3, -2, -1, 1, 2\}$$ contains 5 elements and all matrices are $$3 \times 3$$ with entries from $$S$$, we proceed to count each set.

For $$S_1$$, the symmetric matrices $$A = A^T$$ satisfy $$a_{ij} = a_{ji}$$ so we may choose freely the six entries $$a_{11},\,a_{12},\,a_{13},\,a_{22},\,a_{23},\,a_{33}$$ each having 5 possible values. This gives $$|S_1| = 5^6 = 15625$$.

In the case of skew‐symmetric matrices $$S_2$$ with $$A = -A^T$$ the diagonal entries must satisfy $$a_{ii} = 0$$, but since $$0\notin S$$ no such matrices exist and hence $$|S_2| = 0$$.

Considering $$S_3$$ of trace‐zero matrices where $$a_{11} + a_{22} + a_{33} = 0$$, the six off‐diagonal entries remain free with $$5^6$$ choices. We then count the ordered triples $$(a_{11},a_{22},a_{33}) \in S^3$$ summing to zero. Among the $$5^3 = 125$$ possible triples only $$(-3,1,2)$$ in 6 permutations, $$(-2,1,1)$$ in 3 permutations, and $$(-1,-1,2)$$ in 3 permutations work, giving 12 solutions. Therefore $$|S_3| = 12 \times 5^6 = 187500$$.

Since $$S_2$$ is empty we have $$|S_1 \cap S_2| = 0$$ and $$|S_2 \cap S_3| = 0$$. For $$S_1 \cap S_3$$, symmetric matrices with trace zero allow the three off‐diagonal entries $$a_{12}, a_{13}, a_{23}$$ to be chosen arbitrarily in $$5^3 = 125$$ ways, while the diagonal triple must sum to zero in 12 ways, yielding $$|S_1 \cap S_3| = 12 \times 125 = 1500$$. Moreover, the triple intersection $$|S_1 \cap S_2 \cap S_3|$$ is also 0.

By inclusion-exclusion we obtain $$|S_1 \cup S_2 \cup S_3| = 15625 + 0 + 187500 - 0 - 1500 - 0 + 0 = 201625$$.

Substituting into $$125\alpha = |S_1 \cup S_2 \cup S_3|$$ gives $$125\alpha = 201625$$, which yields $$\alpha = \frac{201625}{125} = 1613$$.

Option X: The answer is 1613.