11. Find all prime numbers \( p, q, r \) that satisfy the equation \( p(p+1) + q(q+1) = r(r+1) \).
11. We divide the solution into two cases: $p=q$ and $p \neq q$, and denote the original equation as ( * ).
(1) When $p=q$, equation (*) becomes: $2 p(p+1)=r(r+1)$.

Clearly, $r$ is an odd prime, so from (1) we know that $2 p \mid r+1$, hence $2 p \leqslant r+1$. Using (1) again, we have $r(r+1)=2 p(p+1) \leqslant (r+1)(p+1)$, thus $r \leqslant p+1$, so $2 p \leqslant r+1 \leqslant p+1+1=p+2$, which means $p \leqslant 2$, thus $p=2$, and at this time $r=3$.
(2) When $p \neq q$, note that $p(p+1)+q(q+1) \leqslant(p+q-1)(p+q)$, so using (*) we have
$r \leqslant p+q-1$.
Rearranging and factoring equation (*), we get $p(p+1)=(r-q)(r+q+1)$.
Considering (2), from (3) we can deduce that $p \mid r+q+1$, and similarly, $q \mid r+p+1$. Since $p, q$ are different primes, then from $p \mid r+p+q+1, q \mid r+p+q+1$ we know
$p q \mid(r+p+q+1)$.
Thus, $p q \leqslant r+p+q+1 \leqslant 2(p+q)$, which means
$(p-2)(q-2) \leqslant 4$.
(4)
Assume $p>q$, by enumeration we find that the pairs $(p, q)$ satisfying (4) are only $(5,3)$, and $q=2, p$ is any odd prime. If $(p, q)=(5,3)$, then $r(r+1)=36$, which is impossible; if $q=2$, from $2 \mid(r+p+1)$ we know $r$ is even, which is also impossible, so there is no solution when $p \neq q$.
The prime triplet $(p, q, r)$ satisfying (*) is $(2,2,3)$.
Throughout this process, we repeatedly used an important technique: if positive integers $a, b$ satisfy $a \mid b$, then $a \leqslant b$.