7-11 Let the four vertices of a regular tetrahedron be $A, B, C, D$, with each edge length being 1 meter. A small insect starts from point $A$ and moves forward according to the following rules: at each vertex, it chooses one of the three edges connected to that vertex with equal probability, and crawls all the way to the end of that edge. Suppose the probability that it is exactly at vertex $A$ after crawling 7 meters is
$$
p=\frac{n}{729} \text {. }
$$

Find the value of $n$.
[Solution 1] Let $a_{n}$ denote the probability that the bug returns to point $A$ after walking $n$ meters $(n=0,1,2, \cdots)$.
If the bug has walked $n-1$ meters and is not at $A$, the probability is
$$
1-a_{n-1},
$$

and the probability of moving from a point outside $A$ to $A$ is $\frac{1}{3}$, so we have
$$
a_{n}=\frac{1}{3}\left(1-a_{n-1}\right).
$$

We also know that $a_{0}=1$, meaning the bug starts from $A$. We can calculate step by step from (1):
$$
\begin{array}{l}
a_{1}=0, a_{2}=\frac{1}{3}, a_{3}=\frac{2}{9}, a_{4}=\frac{7}{27}, \\
a_{5}=\frac{20}{81}, a_{6}=\frac{61}{243}, a_{7}=\frac{182}{729},
\end{array}
$$

From $a_{7}=\frac{n}{729}=\frac{182}{729}$,
we get $\quad n=182$.
[Solution 2] We can also derive the general formula for $a_{n}$ from (1):
$a_{n}-\frac{1}{4}=-\frac{1}{3}\left(a_{n-1}-\frac{1}{4}\right)$, thus
$$
a_{n}-\frac{1}{4}=\left(-\frac{1}{3}\right)^{n}\left(a_{0}-\frac{1}{4}\right) \text {, i.e., } a_{n}-\frac{1}{4}=\frac{3}{4} \cdot\left(-\frac{1}{3}\right)^{n} \text {, }
$$

so $a_{n}=\frac{3}{4} \cdot\left(-\frac{1}{3}\right)^{n}+\frac{1}{4} \quad(n=0,1,2, \cdots)$.
That is, $a_{n}=\frac{1}{4}+\frac{3}{4}\left(-\frac{1}{3}\right)^{n}$,
and we calculate $\quad a_{7}=\frac{182}{729}$.
Thus, $n=182$.