Problem 1375

Let $p, q, r$ be the prime numbers that satisfy the conditions of the problem. If $p > 2$ then all $p, q, r$ are odd thus the number $A = 3p + 5q$ is even and the number $B = (r-p)(r-q)(q-p)+1$ is odd, which contradicts the conditions of the problem. Therefore, $p=2$, thus:

$$(r-2)(r-q)(q-2)+1=6+5q.$$

$p < q < r$, that means $r-2 > q-2$ and $r-q \ge 2$. Therefore, $6+5q > 2(q-2)^2+1$. After solving the last inequality, we will have that $q < 7$. Prime numbers that satisfy this inequality are $q=3$ and $q=5$.

If $q=3$, then $B = 3p+5q = 21$ is not prime.

If $q=5$, then $B = 3p+5q = 31$ is prime, that implies $A = (r-2)(r-5) \cdot 3+1 = 31$, and $r^2-7r=0$, that means $r=7$.