1. Interpret the condition

Let $M=\begin{pmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{pmatrix},$ where each entry is from $\{0,1,2\}$.

We need the sum of diagonal entries of $M^TM$ to be $5$.

Now, $M^TM = \left[\sum_{k=1}^3 a_{k i} a_{k j}\right],$ so the diagonal entries are $(M^TM)_{11} = a_{11}^2+a_{21}^2+a_{31}^2,$ $(M^TM)_{22} = a_{12}^2+a_{22}^2+a_{32}^2,$ $(M^TM)_{33} = a_{13}^2+a_{23}^2+a_{33}^2.$

Hence the sum of diagonal entries is $\operatorname{tr}(M^TM)=\sum_{i=1}^3\sum_{j=1}^3 a_{ij}^2.$

So the condition becomes: $\sum_{i,j} a_{ij}^2 = 5.$

2. Possible squares of entries

Since each entry is in $\{0,1,2\}$, its square is in $\{0,1,4\}$.

We need the sum of the $9$ squares to be $5$.

Let:

• $x$ = number of entries equal to $\pm 1$ — but here entries are only $0,1,2$, so actually $x$ = number of entries equal to $1$,
• $y$ = number of entries equal to $2$.

Then $x+4y=5.$

We need nonnegative integer solutions.

Checking:

• If $y=0$, then $x=5$.
• If $y=1$, then $x=1$.
• If $y\ge 2$, then $4y\ge 8>5$, impossible.

So there are only two cases.

3. Case 1: Five entries are 1, rest are 0

Choose which $5$ of the $9$ positions contain $1$: $\binom{9}{5}=126.$

So this case gives $126$ matrices.

4. Case 2: One entry is 2, one entry is 1, rest are 0

Choose the position of the entry $2$: $9 \text{ choices}.$

Then choose the position of the entry $1$ from the remaining $8$ positions: $8 \text{ choices}.$

Thus total matrices in this case: $9\times 8 = 72.$

5. Total number of matrices

Adding both cases: $126+72=198.$

6. Check with options

The correct option is $\boxed{198}$ which is Option A.

7. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.