Question 12. Let the four vertices of a regular tetrahedron be A, B, C, and D, with each edge length being 1 meter. There is a small insect that 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 \( P = \frac{n}{729} \). Find the value of \( n \).
For $\mathrm{n}=0,1,2, \cdots$, let $\mathrm{a}_{\mathrm{n}}$ denote the probability that the bug returns to point $\mathrm{A}$ after walking $\mathrm{n}$ meters. Then we have $\mathrm{a}_{\mathrm{n}+1} = \frac{1}{3} (1 - a_{n})$ (1), because the necessary and sufficient condition for the bug to reach point $\mathrm{A}$ after walking $\mathrm{n}+1$ meters is: (a) after walking $n$ meters, it is not at point $\mathrm{A}$ (probability is $1 - a_{n}$), and (b) it moves towards $\mathrm{A}$ from a point other than $\mathrm{A}$ (probability is $\frac{1}{3}$). Now $\mathrm{a}_{0}=1$ (i.e., the bug starts from point $\mathrm{A}$), and from equation (1), we can find $a_{1}=a, a_{2}=\frac{1}{3}, a_{3}=\frac{2}{9}, a_{4}=\frac{7}{27}, a_{5}=\frac{20}{81}, a_{6}=\frac{6!}{243}$, and BP = $a_{7} = \frac{182}{729}$. Therefore, $\mathrm{n}=182$.