2. In $\triangle A B C$, $M, N$ are points on line segments $A B, A C$ respectively, and $A M=\frac{1}{2} A B, A N=\frac{1}{3} A C, E, F, G$ are three moving points on line segment $B C$, and $\overrightarrow{A E}+\overrightarrow{A F}+\overrightarrow{A G}=x \overrightarrow{A M}+y \overrightarrow{A N}(x, y \in \mathbf{R})$, then the minimum value of $\frac{2}{x}+\frac{3}{y}$ is $\qquad$
$$
\begin{array}{l}
\text { As shown in the figure, } \overrightarrow{A E}=x_{1} \overrightarrow{A B}+y_{1} \overrightarrow{A C} \Rightarrow x_{1}+y_{1}=1, \\
\overrightarrow{A F}=x_{2} \overrightarrow{A B}+y_{2} \overrightarrow{A C} \Rightarrow x_{2}+y_{2}=1, \\
\overrightarrow{A G}=x_{3} \overrightarrow{A B}+y_{3} \overrightarrow{A C} \Rightarrow x_{3}+y_{3}=1,
\end{array}
$$
$$
\text { then } \overrightarrow{A E}+\overrightarrow{A F}+\overrightarrow{A G}=x \overrightarrow{A M}+y \overrightarrow{A N}
$$
$$
\begin{array}{l}
\Rightarrow\left(x_{1}+x_{2}+x_{3}\right) \overrightarrow{A B}+\left(y_{1}+y_{2}+y_{3}\right) \overrightarrow{A C} \\
=\frac{x}{2} \overrightarrow{A B}+\frac{y}{3} \overrightarrow{A C} \Rightarrow \frac{x}{2}+\frac{y}{3}=3 .
\end{array}
$$

Thus, $\frac{2}{x}+\frac{3}{y}=\frac{1}{3}\left(\frac{x}{2}+\frac{y}{3}\right)\left(\frac{2}{x}+\frac{3}{y}\right) \geqslant \frac{4}{3}$, equality holds when $x=3, y=\frac{9}{2}$. Therefore, the minimum value of $\frac{2}{x}+\frac{3}{y}$ is $\frac{4}{3}$.