We proceed by induction on $n$, with the base case, $n=1$ being clear. For $t=0,1,2$, let $J_t$ be the set of sequences with $x_n=t$. We know that any three sequences in $J_t\cup J_u$ must have $x_j,y_j,z_j$ distinct, but $j\ne n$. First observe $(x_1,\cdots,x_{n-1})$ cannot appear in $J_t$ and $J_u$ simultaneously. By inductive hypothesis, there are at most $2\cdot (\frac 32)^{n-1}$ possibilities for the first $n-1$ elements in $J_t\cup J_u$. Since each of the possibilities can only appear in at most one of $J_t$, $J_u$, $|J_t|+|J_u|\le 2\cdot (\frac 32)^{n-1}$ sequences by inductive hypothesis. Thus, $2(|J_0|+|J_1|+|J_2|) \le 6\cdot (\frac 32)^{n-1}$, completing the induction

Note that total number of unordered triplets $(x,y,z) \in Q^3 _{n} = \frac{6! ^n}{6!} = 2^{n-1}3^{n-1}$ , suppose for the sake of contradiction assume that there exist a good subset with $ > 2 \cdot (\frac{3}{2})^n$ , then note that $\binom{|A|}{3} \le 2^{n-1}3^{n-1}$ , which yields $$(3^n -2^{n-1}) (3^n -2^n) \le 2^{4n-3}$$which is not true for $n \ge 2$. Hence we are done.