Let $$M$$ be any $$3 \times 3$$ matrix with entries from the set $$\{0, 1, 2\}$$. The maximum number of such matrices, for which the sum of diagonal elements of $$M^T M$$ is seven, is ______.

We need to find the number of $$3 \times 3$$ matrices $$M$$ with entries from $$\{0, 1, 2\}$$ such that the sum of diagonal elements of $$M^T M$$ is seven.

The trace of $$M^T M$$ equals the sum of squares of all entries of $$M$$. This is because $$(M^T M)_{ii} = \sum_j M_{ji}^2$$, so $$\text{tr}(M^T M) = \sum_{i,j} M_{ji}^2$$.

Each entry of $$M$$ is $$0$$, $$1$$, or $$2$$, contributing $$0$$, $$1$$, or $$4$$ to the sum of squares, respectively. We need the total sum of squares of all $$9$$ entries to equal $$7$$.

We consider the possible cases.

Case 1: One entry is $$2$$ (contributing $$4$$), three entries are $$1$$ (contributing $$3$$), and five entries are $$0$$. The sum of squares is $$4 + 3 = 7$$. The number of such matrices is $$\binom{9}{1} \times \binom{8}{3} = 9 \times 56 = 504$$.

Case 2: No entry is $$2$$, so seven entries are $$1$$ (contributing $$7$$) and two entries are $$0$$. The number of such matrices is $$\binom{9}{2} = 36$$.

No other combinations of $$0$$, $$1$$, and $$4$$ can sum to $$7$$ (for instance, two entries equal to $$2$$ would already contribute $$8 > 7$$).

The total number of matrices is $$504 + 36 = 540$$.

Hence, the answer is $$540$$.