7. Let $a_{1}, a_{2}, \cdots, a_{6}$ be any permutation of $\{1,2, \cdots, 6\}$. If the sum of any three consecutive numbers cannot be divisible by 3, then the number of such permutations is $\qquad$.
Consider 2 zeros, 2 ones, 2 twos, and any three consecutive numbers cannot sum to a multiple of 3. The valid permutations are: 001122, 002211, 110022, 112200, 220011, 221100, 011220, 022110, 100221, 122001, 200112, 211002. By substituting 1, 2, 3, 4, 5, 6, we get a total of 12 permutations, each multiplied by \(2^3 = 8\), resulting in a total of \(12 \cdot 2^3 = 96\) permutations.