Let $\dot{P} = PR - PS$, $\dot{R} = RS - PR$, and $\dot{S} = PS - RS$. We can easily show that $\frac{d}{dt}( P + R + S) = 0$ and $\frac{d}{dt} (PRS) = 0$.
I'm, however, stumped at how to use this fact to reduce this system of three to a planar system using the fact that $\frac{d}{dt}( P + R + S) = 0$.
So far, I have $$ \dot{S} - \dot{P} - \dot{R} - \dot{S} = PS - RS \implies PS = RS \implies S = 0 \text{ or } P = R.$$
Use $P+R+S=C$, where the constant $C$ is determined by the initial conditions, and use that to eliminate $S$ in the first two equations, $$\dot P=PR-P(C-P-R)=2PR+P^2-PC\\ \dot R = R(C-P-R)-PR = RC -2PR-R^2 $$