12. $A, B, C, D$ are the 4 vertices of a regular tetrahedron, with each edge length being $1 \, \text{m}$. A gecko starts from point $A$ and crawls along the edges, following these rules: it does not change direction midway along any edge, and when it reaches each vertex, it has an equal probability of choosing any of the three edges that meet at that vertex to continue crawling. What is the probability that the gecko, after crawling $7 \, \text{m}$, returns to vertex $A$?
12. $\frac{182}{729}$.

Let the number of ways the gecko can crawl $n \mathrm{~m}$ and return to vertex $A$ be $a_{n}$, and the number of ways to crawl $7 \mathrm{~m}$ and return to vertex $A$ be $a_{1}$. We can derive the recursive relation $a_{n}=3^{n-1}-a_{n-1}, n \geqslant 2$, and it is easy to see that $a_{1}=0$. Then,
$$
\begin{array}{l}
a_{7}=3^{6}-\left(3^{5}-a_{5}\right)=3^{6}-3^{5}+a_{5} \\
=3^{6}-3^{5}+\left(3^{4}-a_{4}\right)=\cdots \\
=3^{6}-3^{5}+3^{4}-3^{3}+3^{2}-3 .
\end{array}
$$

Therefore, $a_{1}=\frac{-3\left(1-3^{6}\right)}{1-(-3)}=\frac{3 \times 728}{4}=3 \times 182$.
Thus, $P(A)=\frac{3 \times 182}{3^{7}}=\frac{182}{729}$.