1. Assume the contrary: Suppose there exists an infinite sequence of triples $(p_i, q_i, r_i)$ of distinct primes such that $p_i > q_i > r_i$ and $p_i \geq p_j$ whenever $i \geq j$.

2. **Behavior of $p_i$**: Since $p_i$ is a sequence of primes that is non-decreasing and unbounded, we have $\lim_{i \to \infty} p_i = +\infty$.

3. **Behavior of $q_i$**: Given $p_i \mid q_i r_i - k$, we have:
$$q_i r_i - k \geq p_i \implies q_i r_i \geq p_i + k$$
Since $q_i$ and $r_i$ are primes and $q_i > r_i$, we can approximate:
$$q_i^2 > q_i r_i \geq p_i + k \implies q_i \geq \sqrt{p_i + k}$$
Therefore, as $p_i \to \infty$, $q_i \to \infty$.

4. **Behavior of $r_i$**: Similarly, since $q_i \to \infty$, and $r_i$ is the smallest prime in the triple, $r_i$ must also tend to infinity as $i \to \infty$.

5. Divisibility condition: From the given conditions:
$$p_i \mid q_i r_i - k, \quad q_i \mid p_i r_i - k, \quad r_i \mid p_i q_i - k$$
We can multiply these conditions:
$$p_i q_i r_i \mid (q_i r_i - k)(p_i r_i - k)(p_i q_i - k)$$

6. Simplification: Expanding the right-hand side, we get:
$$(q_i r_i - k)(p_i r_i - k)(p_i q_i - k) = p_i q_i r_i (q_i r_i + p_i r_i + p_i q_i - 2k) + k^3 - k^2(p_i + q_i + r_i) + k(p_i q_i + q_i r_i + r_i p_i)$$
Since $p_i q_i r_i$ divides the left-hand side, it must also divide the simplified form:
$$p_i q_i r_i \mid p_i q_i r_i (q_i r_i + p_i r_i + p_i q_i - 2k)$$
This implies:
$$p_i q_i r_i \leq p_i q_i r_i (q_i r_i + p_i r_i + p_i q_i - 2k)$$

7. Inequality: Dividing both sides by $p_i q_i r_i$, we get:
$$1 \leq q_i r_i + p_i r_i + p_i q_i - 2k$$
Dividing by $p_i q_i r_i$, we get:
$$1 \leq \frac{1}{p_i} + \frac{1}{q_i} + \frac{1}{r_i} - \frac{2k}{p_i q_i r_i}$$

8. Limit behavior: As $i \to \infty$, $p_i, q_i, r_i \to \infty$, so:
$$\lim_{i \to \infty} \left( \frac{1}{p_i} + \frac{1}{q_i} + \frac{1}{r_i} - \frac{2k}{p_i q_i r_i} \right) = 0$$
This leads to a contradiction since the left-hand side is always 1, but the right-hand side tends to 0.

Therefore, our initial assumption that there exists an infinite sequence of such triples is false. Hence, for each natural number $k$, there exist only finitely many triples $(p, q, r)$ of distinct primes satisfying the given conditions.

$\blacksquare$