Pequliar Primes

Firstly, it is obvious that no two of these primes are equal.Also,

$(pq-1)(pr-1)(qr-1) \equiv pr+pq+qr-1\equiv 0\pmod{pqr}$

We can write that as

$pqr(\frac{1}{q}+\frac{1}{r}+\frac{1}{p})-1\equiv0\pmod{pqr}$

Hence $\frac{1}{q}+\frac{1}{r}+\frac{1}{p}>1$

But it is easily seen that the only prime numbers satisfying the inequality are 2,3 and 5, because $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}<1$

If we check, that triple satisfies the divisibility condition, so the only possible product is $\boxed{30}$

WLOG let us assume $p > q > r$ .Since if $p=q=r$ then $p \nmid p^2-1$ and likewise.Thus the equal to sign is removed.

$p \mid qr -1; q \mid pr-1 ; r \mid pq-1$

$\implies pqr \mid (pq-1)(qr -1)(pr-1)$

$\implies pqr \mid (pq+qr+pr-1)+pqr(pqr-(p+q+r))$ if we expand.Notice that $pqr$ is already a factor of one of the terms.Hence

$\implies pqr \mid (pq+qr+pr-1) \implies pqr \leq pq+qr+pr-1 \implies pqr < pq+qr+pr \implies \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} > 1$

Suppose now that $r \geq 3$ this means $q \geq 5$ and $p \geq 7$ .Thus $\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} \leq \dfrac{71}{105} < 1$ a contradiction.

This forces $r = 2$ .Now suppose $q \geq 5$ thus $p \geq 7$ but this renders $\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{2} \leq \dfrac{59}{70} < 1$ again a contradiction.

Thus the only solution is $\boxed{p = 5 ;q = 3; r = 2}$ and its cyclic permutations.