$21 、$ Arrange the numbers $12, 13, \cdots, 20$ in a row, such that the sum of every three consecutive numbers is a multiple of 3. There are $\qquad$ ways to do this.
【Analysis】Among 9 numbers, exactly 3 numbers are congruent to 0 modulo 3, 3 numbers are congruent to 1 modulo 3, and 3 numbers are congruent to 2 modulo 3.
Since $a_{1}+a_{2}+a_{3} \equiv a_{2}+a_{3}+a_{4}(\bmod 3) \Rightarrow a_{1} \equiv a_{4}(\bmod 3)$
Therefore, $a_{1} \equiv a_{4} \equiv a_{7}(\bmod 3), a_{2} \equiv a_{5} \equiv a_{8}(\bmod 3), a_{3} \equiv a_{6} \equiv a_{9}(\bmod 3)$ and $a_{1}, a_{4}, a_{7}, a_{2}, a_{5}, a_{8}, a_{3}, a_{6}, a_{9}$ can each have a remainder of $0, 1, 2$ chosen in $3 \times 2 \times 1=6$ ways.
Each group has $3 \times 2 \times 1=6$ arrangements.
Thus, there are $6 \times 6 \times 6 \times 6=1296$ arrangements in total.