, 2011

Dividing both sides of the equation by $pqr$, we obtain
$$\left(1 + \frac{1}{p}\right) \left(1 + \frac{2}{q}\right) \left(1 + \frac{3}{r}\right) = 4.$$
If $p, q, r \ge 5$, then
$$\left(1 + \frac{1}{p}\right) \left(1 + \frac{2}{q}\right) \left(1 + \frac{3}{r}\right) \le \frac{6}{5} \cdot \frac{7}{5} \cdot \frac{8}{5} < 4,$$
hence at least one of $p, q$ or $r$ is less than $5$, and since they're all prime, we have the following cases:

Case 1: $p = 2$. Then $3(q+2)(r+3) = 8qr \implies (5q-6)(5r-9) = 144$ which has the solution $(q, r) = (3, 5)$.

Case 2: $p = 3$. Then $4(q+2)(r+3) = 12qr \implies (q-1)(2r-3) = 9$, which has no solutions.

Case 3: $q = 2$. Then $4(p+1)(r+3) = 8pr \implies (p-1)(r-3) = 6$, which has no solutions.

Case 4: $q = 3$. Then $5(p+1)(r+3) = 12pr \implies (7p-5)(7r-15) = 180$ which has the two solutions $(p, r) = (5, 3), (2, 5)$.

Case 5: $r=2$. Then $5(p+1)(q+2) = 8pq \implies (3p-5)(3q-10) = 80$
which has the solution $(p, q) = (7, 5)$.

Case 6: $r = 3$. Then $6(p + 1)(q + 2) = 12pq \implies (p - 1)(q - 2) = 4$
which has the solution $(p, q) = (5, 3)$.

Hence the equation has the three solutions $(p, q, r) = (2, 3, 5), (5, 3, 3)$ and $(7, 5, 2)$.