We are tasked with finding all ordered triples of primes $(p, q, r)$ such that
$$p \mid q^r + 1, \quad q \mid r^p + 1, \quad r \mid p^q + 1.$$

Assume $p = \min(p, q, r)$ and $p \neq 2$. Note the following conditions:
$$\begin{align*} \text{ord}_p(q) &\mid 2r \implies \text{ord}_p(q) = 2 \text{ or } 2r, \\ \text{ord}_q(r) &\mid 2p \implies \text{ord}_q(r) = 2 \text{ or } 2p, \\ \text{ord}_r(p) &\mid 2q \implies \text{ord}_r(p) = 2 \text{ or } 2q. \end{align*}$$

Given that $\text{ord}_p(q) \mid \phi(p) = p-1$ and $\text{ord}_p(q) < r < 2r$, we deduce $\text{ord}_p(q) = 2$, implying $p \mid q + 1$.

Similarly, $\text{ord}_q(r) \mid \gcd(\phi(q) = q-1, 2p) = 2$, so $\text{ord}_q(r) = 2$, implying $q \mid r + 1$.

Repeating this process, we get $r \mid p + 1$. This results in a contradiction unless $p = 2$.

Now, with $p = 2$, we have:
$$q \mid r^2 + 1 \quad \text{and} \quad r \mid 2^q + 1.$$

Assume $\text{ord}_r(2) = 2q$, which implies $2q \mid r - 1$. Thus, $r = 2qk + 1$. Plugging this into $q \mid r^2 + 1$, we find $q \mid 2$, leading to a contradiction. Therefore, $\text{ord}_r(2) = 2$, implying $r = 3$.

Thus, $q = 5$ or $q = 3$. Since all primes must be distinct, we conclude $q = 5$.

Therefore, the ordered triples of primes $(p, q, r)$ satisfying the given conditions are:
$$(p, q, r) = (2, 3, 5), (2, 5, 3), (3, 2, 5), (3, 5, 2), (5, 2, 3), (5, 3, 2).$$

The answer is: \boxed{(2, 3, 5), (2, 5, 3), (3, 2, 5), (3, 5, 2), (5, 2, 3), (5, 3, 2)}.