Final Round

We have $$\frac{p+q}{r}=p-q+r\iff q(r+1)-p(r-1)=r^2.$$ If $r>2$, then $r$ is odd (since $r$ is prime). Thus the left-hand side of the latter equation is an even number while its right-hand side is an odd number, a contradiction. Therefore, $r=2$. Then from the initial equality it follows that $p=3q-4$. So, if $q$ increases, then $p$ also increases. Thus, the product $pqr=(3q-4)\cdot q\cdot 2$ has maximal value as $q$ has maximal value.

Since $p=3q-4$ and, by condition, $p+q<111$, we have $3q-4+q=4q-4<111$, whence $q<29$. Then $q$, as a prime, can admit only the following values $23$, $19$, $17$, $...$, $2$.

If $q=23$, then $p=3q-4=3\cdot 23-4=65$ is a composite number. If $q=19$, then $p=3q-4=3\cdot 19-4=53$ is a prime number.

Thus, the largest possible value of the product $pqr=53\cdot 19\cdot 2=2014$.