National Math Olympiad

The identities $qr=pqr-15p-7pq=p(qr-15-7q)$ imply that $qr$ is divisible by $p$. Since $p$, $q$ and $r$ are prime numbers, there are only two possibilities: $p=q$ or $p=r$.

If $p=q$ we get $15+7q+r=qr$, which can be rewritten as $$22=qr-7q-r+7=(q-1)(r-7).$$ Exactly one of the numbers $q-1$ and $r-7$ is odd, so one of the primes $q$ and $r$ is even. The only possible case is $q=2$ and from this we get $r=29$.

If $p=r$, then we can divide both sides by $r$ to get $15+8q=qr$ or $15=q(r-8)$. Now, $q$ is a prime and divides $15$. We conclude that $q$ is equal to either $3$ or $5$. If $q=3$ then $r=13$, if $q=5$ then $r=11$.

The solutions are $(p,q,r)=(2,2,29)$, $(13,3,13)$ and $(11,5,11)$.