11. Prime numbers $p$, $q$, $r$ satisfy $p+q=r$, and $(r-p) \cdot$ $(q-p)-27 p$ is a perfect square. Then all the triples $(p, q, r)=$ $\qquad$
11. $(2,29,31)$.

From the problem, we know $p<q<r$, combined with $p+q=r$, we know $p=$ 2. Therefore, $q=r-2$. Thus,
$$
(r-p)(q-p)-27 p=(q-1)^{2}-55 \text {. }
$$

Let $(q-1)^{2}-55=n^{2}$ (where $n$ is a positive integer). Then
$$
(q-1-n)(q-1+n)=55 \text {. }
$$

Solving this, we get $(q, n)=(29,27)$ or $(9,3)$ (discard the latter).
Therefore, $(p, q, r)=(2,29,31)$.