For which values of $p$ and $q$ integers strictly positive does $pq$ divide $3(p-1)(q-1) ?$
It is necessary (but not sufficient) that $p$ and $q$ both divide $3(p-1)(q-1)$. Since $p$ is coprime with $p-1$, by Gauss's theorem, $p$ must divide $3(q-1)$. Similarly, $q$, being coprime with $q-1$, must divide $3(p-1)$. Suppose that $p \geq q$ (since $p$ and $q$ play symmetric roles, we can assume this "without loss of generality"). $p$ divides $3(q-1) \leq 3(p-1) < 3p$, so we can have: either $3(q-1) = 2p$, or $3(q-1) = p$. In the first case, $q$ must divide $3(p-1) = 3\left(\frac{3(q-1)}{2} - 1\right) = \frac{9q - 15}{2}$, so $q$ must divide $9q - 15$, and thus must divide 15, which offers four possibilities:

$q = 15$, hence $p = \frac{3(q-1)}{2} = 21$, but $21 \times 15$ does not divide $3 \times 20 \times 14$ although 21 divides $3 \times 14$ and 15 divides $3 \times 20$.

$q = 5, p = 6$ which is indeed a solution: $6 \times 5$ divides $3 \times 5 \times 4$.

$q = 3, p = 3$ is not a solution, and $q = 1, p = 0$ is not either.

If now $3(q-1) = p$, $q$ must divide $3(p-1) = 3(3(q-1) - 1) = 9q - 12$, so $q$ must divide 12: $q = 12, p = 3(q-1) = 33$ is not a solution of the original equation, $q = 6, p = 15$ is not either, but $q = 4, p = 9$ is a solution, $q = 3, p = 6$ is not, $q = 2, p = 3$ is still a solution, $q = 1, p = 0$ is not. The problem therefore has three solutions, up to permutation: $(6,5)$, $(4,9)$, and $(2,3)$.