We need to find $n(S_1\cup S_2\cup S_3).$ Given that $n(S_1\cup S_2\cup S_3)=125\alpha,$ we must compute $\alpha$.

1. Count $|S_1|$

$S_1$ is the set of all symmetric $3\times 3$ matrices with entries from $S=\{-3,-2,-1,1,2\}.$

A symmetric $3\times 3$ matrix has the form

So there are $6$ independent entries.

Each independent entry can be chosen in $5$ ways. Hence, $|S_1|=5^6=15625.$

2. Count $|S_2|$

$S_2$ is the set of all skew-symmetric matrices with entries from $S$.

A skew-symmetric matrix satisfies $A=-A^T,$ so diagonal entries must satisfy $a_{ii}=-a_{ii}\implies a_{ii}=0.$

But $0\notin S$. Hence no such matrix exists. Therefore, $|S_2|=0.$

3. Count $|S_3|$

$S_3$ consists of all $3\times 3$ matrices with entries from $S$ such that $a_{11}+a_{22}+a_{33}=0.$

Step 3.1: Count choices for diagonal entries

We need ordered triples $(x,y,z)\in S^3$ such that $x+y+z=0.$

Since all entries are from $S=\{-3,-2,-1,1,2\},$ we count all possible ordered triples.

Possible value combinations summing to $0$:

1. $(-3,1,2)$ and its permutations
2. $(-2,1,1)$ and its permutations
3. $(-1,-1,2)$ and its permutations

Now count ordered triples:

• For $(-3,1,2)$: all distinct, so number of permutations is $3!=6.$
• For $(-2,1,1)$: two 1's repeated, so number of permutations is $\frac{3!}{2!}=3.$
• For $(-1,-1,2)$: two $-1$'s repeated, so number of permutations is $\frac{3!}{2!}=3.$

Thus total number of diagonal triples is $6+3+3=12.$

Step 3.2: Count off-diagonal entries

The remaining $6$ entries of a $3\times 3$ matrix are unrestricted, each with $5$ choices. So number of choices is $5^6.$

Hence, $|S_3|=12\cdot 5^6=12\cdot 15625=187500.$

4. Count intersections

We use inclusion-exclusion.

4.1: $S_1\cap S_2$

A matrix both symmetric and skew-symmetric satisfies $A=A^T \quad\text{and}\quad A=-A^T.$ So $A=-A \implies 2A=0 \implies A=0.$ But the zero matrix is not allowed since $0\notin S$. Hence, $|S_1\cap S_2|=0.$

4.2: $S_2\cap S_3$

Since $S_2=\varnothing$, $|S_2\cap S_3|=0.$

4.3: $S_1\cap S_3$

These are symmetric matrices whose diagonal entries sum to $0$.

A symmetric matrix has $6$ independent entries. Among the $3$ diagonal entries, the ordered triple must sum to $0$, and we already counted such triples: $12$ ways. The $3$ independent off-diagonal entries can each be chosen in $5$ ways.

Therefore, $|S_1\cap S_3|=12\cdot 5^3=12\cdot 125=1500.$

4.4: Triple intersection

Since $S_2=\varnothing$, $|S_1\cap S_2\cap S_3|=0.$

5. Apply inclusion-exclusion

Substitute values:

So,

Given $201625=125\alpha,$ we get

6. Final Answer

$\boxed{1613}$

The derived answer matches the stored correct answer.