Czech-Slovak-Polish Match

(Mutually distinct) primes $p,q,r$ satisfy the desired conditions if and only if the number $pq+pr+qr-k$ is divisible by each of the primes $p,q,r$; that is, by the product $pqr$. The equality $pq+pr+qr-k=n\cdot pqr$, for a suitable integer $n$, can be rewritten as $k=pq+pr+qr-n\cdot pqr$. If $n\le 0$, then the last equality implies that $\max \{pq,pr,qr\}\le k$; however, then each of the primes $p,q,r$ is less than or equal to $k/2$ (and there is only a finite number of such triples).

If $n\ge 1$, then we get the estimate $k\le pq+pr+qr-pqr$. Let us show that the last expression is negative (contradicting the fact that $k>0$) unless the triple in question is $\{p,q,r\}=\{2,3,5\}$. We can assume that $2\le p<q<r$ and $r\ge 7$. Then $pq\ge 2\cdot 3=6$ and the inequality $(p-2)(q-2)\ge 0$ implies that $p+q\le \frac{1}{2}pq+2$, hence $$\begin{gathered} pq+pr+qr-pqr=(p+q)r+pq-pqr\le (\frac{1}{2}pq+2)r+pq-pqr \\ =2r-pq(\frac{1}{2}r-1)\le 2r-6(\frac{1}{2}r-1)=6-r<0. \end{gathered}$$