Example 11 (55th Czech and Slovak Mathematical Olympiad) Find all triples of distinct primes $(p, q, r)$ such that $p|(q+r), q|(r+2p), r \mid(p+3q)$.
(1) If $p$ is the largest prime among $p, q, r$, from $p \mid (q+r)$ and $q+r < 2p$, we get $q+r = p$. From $q \mid (r+2p)$, we get $q \mid (3r+2q)$, i.e., $q \mid 3r$.
Since $q \neq r$, we have $q = 3$.
Thus, $p = r + 3$.
From $r \mid (p+3q)$, we get $r \mid (r+12)$, i.e., $r \mid 12$, so $r = 2, p = 5$.
(2) If $q$ is the largest prime among $p, q, r$, since $q \mid (r+2p)$ and $r+2p < 3q$, then $r+2p = q$ or $r+2p = 2q$.
If $2q = r+2p$, then $r$ is even, thus $r = 2$. Therefore, $q = p+1$, which is a contradiction.
If $q = r+2p$, from $p \mid (q+r)$, we get $p \mid (2r+2p)$, i.e., $p \mid 2r$. Therefore, $p = 2$.
From $r \mid (p+3q)$, we know $r \mid (3r+7p)$, so $r \mid (3r+14)$, i.e., $r \mid 14$, so,
$r = 7, q = 11$.
(3) If $r$ is the largest prime among $p, q, r$, since $r \mid (p+3q)$ and $p+3q < 4r$, then $p+3q = 3r$ or $p+3q = 2r$ or $p+3q = r$.
If $p+3q = 3r$, then $p = 3$. Therefore, $r = q+1$, which is a contradiction.
If $p+3q = 2r$, from $p \mid (q+r)$, we get $p \mid 2(q+r)$, so $p \mid (p+5q)$, i.e., $p \mid 5q$. Therefore, $p = 5$.
From $q \mid 2(r+2p)$, we get $q \mid (3q+25)$, i.e., $q \mid 25$, so, $q = 5$, which is a contradiction.
If $p+3q = r$, from $p \mid (q+r)$, we get $p \mid (p+4q)$, i.e., $p \mid 4q$, so, $p = 2$.
From $q \mid (r+2p)$, we get $q \mid (3q+6)$, i.e., $q \mid 6$, so,
$q = 3, r = p+3q = 11$.
In summary, there are three sets of numbers that satisfy the conditions:
$$
(5,3,2),(2,11,7),(2,3,11) \text{. }
$$