If $p,q,r$ is distinct,$pqr \mid pq(p+q)+qr(q+r)+rp(r+p)$, then $p \mid q+r$, $q \mid r+p$, $r \mid p+q$, $gcd(p,q)=gcd(q,r)=gcd(r,p)=1$, we can get $pqr \mid p+q+r$, then it's easy.

Finished solution: Suppose p,q,r are distinct.We have pqr <=p+q+r and pqr>2qr>4r, pqr>4p, pqr>4q.Hence pqr>4/3 (p+q+r)>p+q+r, a contradiction.Therefore p=q=r, a=6.