Let $X$ be the intersection of $\overline{CP}$ and $\overline{AB}$. 

Since $\overline{PQ} \parallel \overline{CA}$ and $\overline{PR} \parallel \overline{CB}$, $\angle CAB = \angle PQR$ and $\angle CBA = \angle PRQ$. So $\Delta ABC \sim \Delta QRP$, and thus, $\frac{[\Delta PQR]}{[\Delta ABC]} = \left(\frac{PX}{CX}\right)^2$. 
Using [mass points](https://artofproblemsolving.com/wiki/index.php/Mass_points): 
[WLOG](https://artofproblemsolving.com/wiki/index.php/WLOG), let $W_C=15$. 
Then: 
$W_A = \left(\frac{CE}{AE}\right)W_C = \frac{1}{3}\cdot15=5$.
$W_B = \left(\frac{CD}{BD}\right)W_C = \frac{2}{5}\cdot15=6$. 
$W_X=W_A+W_B=5+6=11$. 
$W_P=W_C+W_X=15+11=26$. 
Thus, $\frac{PX}{CX}=\frac{W_C}{W_P}=\frac{15}{26}$. Therefore, $\frac{[\Delta PQR]}{[\Delta ABC]} = \left( \frac{15}{26} \right)^2 = \frac{225}{676}$, and $m+n=\boxed{901}$. Note we can just use mass points to get  $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ which is $\boxed{901}$.
In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.