a finite group $G$ satisfies that $|G|=pqr$, where $p<q<r$ are primes. Show that $n_r(G)=1$, where $n_r$ is the number of Sylow-r subgroups of G.

now I know $n_q$ must be $1$ because according to sylow theorems, $n_r$ could be $1, pq$. if it is $pq$, then: $n_q$ could be $1,r$, if it's $r$, we have $1+(q-1)r+(r-1)pq=pqr+qr-r-pq+1>pqr$, LHS is the size of all Sylow-q and Sylow-r subgroups, so there is contradiction.so must be $n_q=1$.

However i am stuck for further work. I know next I should consider $Q \in Syl_q(G),~Q\lhd G$ and consider $|G/Q|=pr$.