(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
$$pq + pr + qr - pqr = (p+q)r + pq - pqr \le (\frac{1}{2}pq + 2)r + pq - pqr \\ = 2r - pq(\frac{1}{2}r - 1) \le 2r - 6(\frac{1}{2}r - 1) = 6 - r < 0.$$