Chameleons Riddle

Let $R$, $B$, and $G$ be the number of red, blue, and green chameleons at the moment. After any meeting of chameleons, we have one of $$R\mapsto R,\quad B\mapsto B,\quad G\mapsto G$$

$$R\mapsto R-1,\quad B\mapsto B-1,\quad G\mapsto G+2$$ $$R\mapsto R+2,\quad B\mapsto B-1,\quad G\mapsto G-1$$ $$R\mapsto R-1,\quad B\mapsto B+2,\quad G\mapsto G-1$$ Note that $B-G\pmod 3$, $R-B\pmod 3$, and $G-R\pmod 3$ are preserved under any meeting. Supposing (WLOG) that all chameleons at some point became green, then both $R$ and $B$ would be $0$, and thus $R-B\equiv 0\pmod 3$. But having started with $R-B\equiv 2\pmod 3$, this is impossible. Thus, the chameleons can never all be the same color.

Let the numbers of red, blue, and green chameleons mod $3$ be $\langle r,b,g\rangle$. When two of different colors meet, the resulting numbers after the color changes are $\langle r-1,b-1,g+2\rangle$, $\langle r-1,b+2,g-1\rangle$, or $\langle r+2,b-1,g-1\rangle$. Each of these is the same mod $3$ as a change to $\langle r-1,b-1,g-1\rangle$. Since the initial numbers are $\langle 1,2,0\rangle$, all of which are distinct, and they will always be distinct. In fact, they cycle through the permutations $\langle 1,2,0\rangle$, $\langle 0,1,2\rangle$, and $\langle 2,0,1\rangle$.

Here is a more mechanical, far less elegant strategy, which ended up being the same as above. I realize that one is looking for invariants, but always have a hard time spotting them. The following technique may provide some solution structure who share my difficulty...

Let $A = \begin{bmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end{bmatrix}$. I look for 'neat' eigenvalue/left eigenvector pairs, and find $v^T=(1,-1,0)$ corresponding to eigenvalue $3$. The 'system' we have is $x_{x+1} = x_n + u_n$, where at each $n$, $u_n$ is one of the three columns of $A$, and $x_n = (r_n, g_n, b_n)^T$ represents the number of each color at time $n$. The solution is $x_n = x_0+u_0+...+u_{n-1}$.

Now I look at $v^T x_n$, and notice that $v^T x_n \pmod 3 = v^T x_0 \pmod 3 = 1$. However, $v^T (33,0,0) = 0 \pmod 3$, and similarly for the other extreme distributions. Hence we can never reach these states.