= Solution
Use the <Sylow theorems>. The number $n_p$ divides $qr$ and is congruent to one modulo $p$. Since $q,r<p$, it is either one or $qr$. Suppose $n_p=qr$. Different subgroups of prime order intersect trivially, so their nonidentity elements occupy $qr(p-1)$ places, leaving only $qr-1$ nonidentity elements of other prime orders.
If neither the Sylow $q$-subgroup nor the Sylow $r$-subgroup is normal, then $n_q\geq p$ and $n_r\geq q$. Indeed $n_q$ divides $pr$, and its possible divisor $r<q$ cannot satisfy the Sylow congruence; the smallest remaining nontrivial possibility is at least $p$. Likewise any nontrivial divisor of $pq$ is at least $q$. Their elements would require at least
$$
p(q-1)+q(r-1)>qr-1
$$
places, since the excess is $(p-1)(q-1)>0$. Hence some <Sylow subgroup> of order $s\in\{q,r\}$ is normal.
In the quotient by this <normal subgroup>, the largest prime $p$ has a normal <Sylow subgroup>: for a group of order $pt$ with $p>t$ its Sylow count divides $t<p$ and so equals one. Pulling back gives a <normal subgroup> of order $ps$. Inside it, the subgroup of order $p$ is again the unique Sylow $p$-subgroup. It is characteristic in that <normal subgroup> and therefore normal in $G$, contradicting $n_p=qr$. Thus $\boxed{n_p=1}$.
Let $P$ be this normal <Sylow subgroup>. In $G/P$, of order $qr$, its subgroup of order $q$ is normal by the same argument. Its preimage $K$ is a \b[normal Hall $\{p,q\}$-subgroup]. The series
$$
1\lhd P\lhd K\lhd G
$$
has factors of orders $p,q,r$, hence cyclic and abelian. Therefore $\boxed{G\text{ is soluble}.}$ This establishes solubility before using any Hall-existence conclusion that itself assumes solubility.
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