2. 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 $P$, draw $PQ \parallel CA$ and $PR \parallel CB$, which intersect side $AB$ at points $Q$ and $R$, respectively. The ratio of the area of $\triangle PQR$ to the area of $\triangle ABC$ is $\qquad$
Draw $E F / / A D$ intersecting $B C$ at point $F$. Then
$$
\begin{array}{l}
\frac{C F}{F D}=\frac{C E}{E A}=\frac{2}{5} \\
\Rightarrow F D=\frac{5}{7} . \\
\text { Also, } P Q / / C A, \text { so } \\
\frac{P Q}{E A}=\frac{B P}{B E}=\frac{B D}{B F}=\frac{28}{33} \Rightarrow P Q=\frac{140}{33} .
\end{array}
$$

Also, $P Q / / C A$, so

From $\triangle P Q R \backsim \triangle C A B$, we get
$$
\frac{S_{\triangle P Q R}}{S_{\triangle A B C}}=\left(\frac{P Q}{C A}\right)^{2}=\frac{400}{1089} .
$$