Solution. It is obvious that one of the numbers $\mathbf{p}, \mathbf{q}$ and r is 3. Without loss of generality, let $p=3$. Then $q \mathbf{q}=3+q+r$, from which it follows that $\mathrm{q}(\mathrm{r}-1)-(\mathrm{r}-1)=4$ i.e. $(\mathrm{q}-1) \cdot(\mathrm{r}-1)=4=1 \cdot 4=4 \cdot 1=2 \cdot 2$. Then we get three systems of equations with two unknowns $\left\{\begin{array}{l}q-1=1 \\ r-1=4\end{array} ;\left\{\begin{array}{l}q-1=4 \\ r-1=1\end{array}\right.\right.$ and $\left\{\begin{array}{l}q-1=2 \\ r-1=2\end{array}\right.$ whose solutions are $(2,5),(5,2)$ and $(3,3)$ respectively. The required numbers are $(3,2,5),(3,5,2)$ and $(3,3,3)$.