Assume WLOG that $p \geq q \geq r$. Since $p$ divides $q-r$, it follows $q=r$ because of size. Then $q$ divides $r+n$ implies $q$ divides $n$. But we have $q$ divides $p+n$, too, so we must in fact have $q \mid p$. Since $p$ is prime, we must have $p=q=r$.

Cute Problem Without loss of generality, suppose that $p \geq q \geq r$. We may rewrite the given condition using divisibility, $$pq \mid (r+n),\ qr \mid (p+n),\ rp \mid (q+n).$$Then, we get $$p \mid (q+n),\ p \mid (r+n) \implies p \mid q-r$$and this implies $q=r$, since $0 \leq q-r < p$. Then, we get $$pq \mid (r+n) \implies q \mid (q+n) \implies q \mid n$$and $$q \mid (p+n) \implies q \mid p \implies p=q$$which means $p=q=r$, as desired. $\blacksquare$

WLOG that $p \ge q \ge r$. We have \[n \equiv -p \pmod{qr}, n \equiv -q \pmod{pr}, n \equiv -r \pmod{pq} \] This implies that $r\mid n + p, r\mid n + q$, so $r \mid p -q$. Similarly, we have $p \mid q - r$ and $q \mid p - r$. Since $p > |q - r|$ and $p \mid q - r$, we must have $q - r = 0$, so $q = r$. Now $q\mid p - r = p - q$, so $q\mid p\implies q= p $. Hence $p = q = r$.