Every $\boxed{ }\boxed{ }\boxed{ }$ covers $1,2,3$, so the $\boxed{ }$ must covers $1$.

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And after rotation the $1$ must be an other $1$, so the $\boxed{ }$ must cover a red squre. It is easy to see we can do it.

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First do the coloring $123123\cdots 12$ followed by $231231\cdots 23$ and $312312\cdots 31$. The number of ones, twos, and threes in first nine rows are balanced. Focusing therefore on the last two rows, we find that there are $8$ twos and $7$ of each ones and threes. Hence, with this coloring, $\boxed{ }$ should occupy a squared labeled with a two (since $\boxed{}\boxed{}\boxed{}$ occupies one of each colors). These are blue in my figure below. Next, alternate the coloring as follows. As above, start again with $123123\cdots 12$. But now, follow it with $312312\cdots 31$ and then $231231\cdots 23$. Focusing, as above, again on the last two rows, we now find that the number of ones is one greater than that of twos and threes. Hence, $\boxed{}$ should occupy a one. These are red squares in my drawing. Finally, dash those squares that obtained both a blue as well as a red color. Those are precisely the ones that a $\boxed{}$ should occupy. A construction is immediate, see again my attachment.

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