12. Let $a_{i}=3 i-2, b_{i}=3 i-1, i=1,2, \cdots, n$. Then for any $k \in \mathbf{N}^{*}, k<n$, we have
$$a_{i}+a_{i+1} \equiv 2(\bmod 3), a_{i}+b_{i} \equiv 0(\bmod 3), b_{i}+b_{i+k} \equiv 1(\bmod 3)$$

Thus, when we denote
$$A=\left\{a_{i}+a_{i+1} \mid 1 \leqslant i \leqslant n\right\}, B=\left\{a_{i}+b_{i} \mid 1 \leqslant i \leqslant n\right\}, C=\left\{b_{i}+b_{i+k} \mid 1 \leqslant\right.$$
$i \leqslant n\}$,
in the sets $A, B, C$, any two numbers taken from different sets are not congruent modulo $3 n$ (since they are not congruent modulo 3). Therefore, to prove that $A \cup B \cup C$ forms a complete residue system modulo $3 n$, it suffices to show that any two numbers in $A$ are not congruent modulo $3 n$ (the same can be shown for $B$ and $C$).

If $a_{i}+a_{i+1} \equiv a_{j}+a_{j+1}(\bmod 3 n)$, then
$$2 a_{i}+3 \equiv 2 a_{j}+3(\bmod 3 n)$$

This leads to $a_{i} \equiv a_{j}(\bmod 3 n)$, and thus $i=j$. This shows that any two numbers in $A$ are not congruent modulo $3 n$.