1. Interpret the condition

Let

be a $3\times 3$ matrix whose entries are from $\{0,1,2\}$.

We are given that the sum of diagonal elements of $M^T M$ is $7$.

Now,

because the diagonal entries of $M^T M$ are the sums of squares of the corresponding columns of $M$.

So the condition becomes:

where each $m_{ij} \in \{0,1,2\}$.

2. Translate into a counting problem

Each entry contributes:

• $0^2 = 0$
• $1^2 = 1$
• $2^2 = 4$

We need the sum of squares of the $9$ entries to be $7$.

Suppose:

• number of entries equal to $2$ is $a$
• number of entries equal to $1$ is $b$
• number of entries equal to $0$ is $9-a-b$

Then

with $a,b \ge 0$ integers.

3. Find all possible distributions

Solve

Possible values of $a$:

• If $a=0$, then $b=7$
• If $a=1$, then $b=3$
• If $a\ge 2$, then $4a \ge 8$, impossible

So only two cases are possible:

Case 1: $a=0, b=7$

That means:

• seven entries are $1$
• two entries are $0$
• no entry is $2$

Number of such matrices:

Case 2: $a=1, b=3$

That means:

• one entry is $2$
• three entries are $1$
• five entries are $0$

Number of such matrices:

• choose position of the single $2$: $\binom{9}{1}=9$
• from remaining $8$ positions, choose $3$ positions for the $1$'s: $\binom{8}{3}=56$

Hence number of matrices in this case:

4. Total number of matrices

Therefore total number is

5. Final answer

The required number of matrices is

6. Comparison with stored answer

Stored correct answer = $540$.

Our derived answer is also $540$, so they agree.