8. Let $Q$ be the midpoint of side $BC$ of $\triangle ABC$, $P$ be a point on $AC$ such that $CP=3PA$, and $R$ be a point on $AB$ such that $S_{\triangle PQR}=2S_{\triangle RBQ}$. If $S_{\triangle ABC}=300$, find $S_{\triangle PQR}$.
8. As shown in Figure 3, connect $A Q$ and $C R$.
From 300
$$
\begin{array}{l}
=S_{\triangle A P R}+S_{\triangle P Q R}+S_{\triangle B Q R}+S_{\triangle C Q P}, \\
S_{\triangle B Q R}=\frac{1}{2} S_{\triangle P Q R}, \\
S_{\triangle A P R}=\frac{1}{4} S_{\triangle A R C}=\frac{1}{4}\left(S_{\triangle A B C}-2 S_{\triangle B Q R}\right) \\
\quad=\frac{1}{4}\left(300-S_{\triangle P Q R}\right), \\
S_{\triangle C Q P}=\frac{3}{4} S_{\triangle A Q C}=\frac{3}{8} S_{\triangle A B C}=\frac{3}{8} \times 300,
\end{array}
$$

we get $S_{\triangle P Q R}=90$.