Concurrent Cevians

If the areas of $APS$ , $BPQ$ and $CPR$ are $x,y,z$ respectively, the Ceva's Theorem tells us that $\frac{x}{168} \times \frac{z}{14} \times \frac{y}{90} \; = \; 1$ so that $xyz= 211680$ . Comparing the areas of $APS$ , $AQS$ , $APR$ and $AQR$ tells us that $\frac{x}{168} \; = \; \frac{x+y+90}{z+182}$ and, similarly $\frac{z}{14} \; = \; \frac{x+z+168}{y+104} \hspace{2cm} \frac{y}{90} \; = \; \frac{y+z+14}{x+258}$ Solving these various identities gives only one positive solution for $x,y,z$ , namely $x=252$ , $y=15$ , $z=56$ . This makes the area of the triangle $QRS$ equal to $\boxed{595}$ .