= Solution
Use the $\mathfrak{sl}_3$ <weight of a representation> convention from Question 1: $L_1+L_2+L_3=0$, with <simple roots> $L_1-L_2$ and $L_2-L_3$. This identification is necessary: the general preceding setup does not by itself specify which <root system> the symbols $L_i$ refer to.
The <Weyl group> is $S_3$, acting by permutations of the $L_i$. Let $s_1=(12)$ and $s_2=(23)$, with composition read from right to left. Then
$$
-2L_1+L_2\xrightarrow{s_1}L_1-2L_2\xrightarrow{s_2}L_1-2L_3.
$$
Hence
$$
\boxed{w=s_2s_1=(1\ 3\ 2),\qquad\mu=L_1-2L_3.}
$$
For direct verification, the diagonal coefficients of $\mu$ are $(1,0,-2)$, so its two <Dynkin labels> are $1-0=1$ and $0-(-2)=2$, both strictly positive. Thus $\mu$ is in the open <Weyl chamber>. The original three coefficients $(-2,1,0)$ are distinct, so its stabilizer is trivial and this permutation is unique.
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