9. Given that $D$ and $E$ are points on the sides $BC$ and $CA$ of $\triangle ABC$, respectively, and $BD=4$, $DC=1$, $AE=5$, $EC=2$. Connecting $AD$ and $BE$, they intersect at point $P$. Through point $P$, draw $PQ \parallel CA$ and $PR \parallel CB$, which intersect side $AB$ at points $Q$ and $R$, respectively. Then the ratio of the area of $\triangle PQR$ to the area of $\triangle ABC$ is $\qquad$
9. $\frac{400}{1089}$.

As shown in Figure 6, draw $E F / / A D$, intersecting side $B C$ at point $F$. Then,
$$
\frac{C F}{F D}=\frac{C E}{E A}=\frac{2}{5} .
$$

Thus, $F D=\frac{5}{7}$.
Since $P Q / / C A$, we have,
$$
\frac{P Q}{E A}=\frac{B P}{B E}=\frac{B D}{B F}=\frac{28}{33} \text {. }
$$

Therefore, $P Q=\frac{140}{33}$.
Since $P Q / / C A$ and $P R / / C B$, $\triangle P Q R \backsim \triangle C A B$. Hence, $\frac{S_{\triangle P Q R}}{S_{\triangle C A B}}=\left(\frac{P Q}{C A}\right)^{2}=\left(\frac{20}{33}\right)^{2}=\frac{400}{1089}$.