Label the common interior point $S$, the intersection opposite $P$, $T$, opposite $Q$, $U$ and opposite $R$, $V$. Let $\triangle{PSU}$ be $x$ and $\triangle{RST}$ be $y$.$$\text{ }$$The bases of two adjacent triangles sharing a common side and having the same height are in the same ratio as the ratio of their areas, hence$$\frac34(124+x)=65+y\Rightarrow112=4y-3x$$and$$\frac12(84+x)=y\Rightarrow84=2y-x$$Solving this system of equations yields $x=56$ and $y=70$, so the area of $\triangle{PQR}$ is $315\text{ units}^2$.