It suffices to show that the case $n = 13$ is impossible since we can remove extra points. For the $j$th point, let $r_j, y_j, g_j$ be the numbers of lines having this point as an endpoint which are in red, yellow, and green respectively.

For any $\triangle XYZ$, WLOG assume $XY$ and $XZ$ are red. Then this triangle corresponds to two red lines from $X$, and this correspondence is one-to-one. It follows that the number of triangles is
$$\sum_{j=1}^{13} \left[ \binom{r_j}{2} + \binom{y_j}{2} + \binom{g_j}{2} \right].$$
But then it is clear that there are $\binom{13}{3} = 286$ triangles. By the pigeonhole principle, WLOG assume
$$\binom{r_1}{2} + \binom{y_1}{2} + \binom{g_1}{2} \ge \frac{286}{13} = 22.$$
This gives $r_1(r_1 - 1) + y_1(y_1 - 1) + g_1(g_1 - 1) \ge 44$. Since $r_1 + y_1 + g_1 = 12$, this implies $r_1^2 + y_1^2 + g_1^2 \ge 56$.
We claim that one of $r_1, y_1, g_1$ is at least 6. Suppose on the contrary that $r_1, y_1, g_1 \le 5$. As $r_1+y_1+g_1 = 12$, it suffices to check $(r_1, y_1, g_1) = (5, 5, 2), (5, 4, 3), (4, 4, 4)$. In all cases, $r_1^2 + y_1^2 + g_1^2 \ge 56$ does not hold.
WLOG assume $r_1 \ge 6$. Suppose $AB, AC, AD, AE, AF, AG$ are red. By assumption, none of the lines formed by $B, C, D, E, F, G$ is red. Since $R(3,3) = 6$, there must be a yellow or green triangle among these 6 points, contradiction. This proves $n < 13$.