The given congruences imply $$ pqr\mid pq+qr+rp-1 $$ This is because $p,q,r$ primes means that it suffices to verify that each of $p,q,$ and $r$ divide $pq+qr+rp-1$, but this is clear. Now clearly $pq+qr+rp-1>0$, so then we have $pqr\le pq+qr+rp-1$. WLOG assume that $p\ge q\ge r\ge 2$. Then $$ q(qr-q-r)\le p(qr-q-r)\le qr-1\Rightarrow q^2(r-1)-2qr\le-1 $$ But then $$ qr(\frac{q}{2}-2)\le\frac{q^2r}{2}-2qr\le q^2(r-1)-2qr<0 $$ So $q\le 3$.
If $q=2$, then $p=2$, so $4r\mid 4r+3$, which is impossible for $r\ge 2$.
If $q=3$ and $p=2$, then $6r\mid 5(r+1)$, so $r\mid 5$ give $r=5$ as the only possibility.
If $q=3$ and $p=3$, then $9r\mid 6r+8$, so $r\mid 2$, a contradiction.