BE * $S$ is the intersection point of the lines $l_{1}: 7 x+5 y+8=0$ and $l_{2}: 3 x+4 y-13=0$. Points $P(3, 7)$, $Q(11,13)$ form the line $P Q$ on which there are two points $A$ and $B$. Point $P$ is between $A$ and $Q$, and point $B$ is between $P$ and $Q$, with $\frac{|P A|}{|A Q|}=\frac{|P B|}{|B Q|}=\frac{2}{3}$. Without finding the coordinates of $S$, try to find the equations of the lines $S A$ and $S B$.
From the problem, we have the equation of $S A$ as: $(7 x-5 y+8)+\lambda(3 x+4 y-13)=0$,
(1) The equation of $S B$ as: $(7 x-5 y+8)+\mu(3 x+4 y-13)-0$.
(2) From $\frac{|P A|}{|A Q|}=$ $\frac{|P B|}{|B Q|}=\frac{2}{3}$ and the section formula, we get the coordinates of $A, B$ as $A(-13,-5), B\left(\frac{31}{5}, \frac{47}{5}\right)$. Substituting these into (1) and (2), we find $\lambda=-\frac{29}{36}, \mu=-\frac{11}{108}$. Therefore, the equations of the lines $S A$ and $S B$ are: $S A$ equation: $165 x-296 y+665=0, S B$ equation: $723 x-584 y+1007=0$.