By simple bounding we eliminate $p+q+r=101pqr$. Now we consider $pqr=101(p+q+r)$. Since $101|pqr$, one of the primes has to be 101. WLOG $p=101$. Now we have $qr=101+q+r$, or $(q-1)(r-1)=102$. So the prime triplet is $(2,101,103)$ or a permutation. Hence proved.

$p+q+r=101pqr$ is imposible because $101pqr>pqr>p+q+r$ So $101p+101q+101r=pqr$ WLOG assume $p=101$ $101+q+r=qr$ $qr-q-r-101=0$ $qr-q-r+1=102$ $(q-1)(r-1)=102$ By testing every possible value, get that only $p,r=2,103$ work So the primes are $\boxed{2,101,103}$