11 Prime numbers $p$, $q$, $r$ satisfy $p+q=r$, and $(r-p)(q-p)-27 p$ is a perfect square, then all triples $(p, q, r)=$ $\qquad$ .
11 From the given conditions, we know $p<q<r$, combined with $p+q=r$, we know $p=2$. So $q=r-2$, then
$$
\begin{aligned}
(r-p)(q-p)-27 p & =q(q-2)-54 \\
& =(q-1)^{2}-55 .
\end{aligned}
$$

Let $(q-1)^{2}-55=n^{2}$ ( $n$ is a positive integer), then
$$
(q-1-n)(q-1+n)=55,
$$

we have
$$
\left\{\begin{array}{l}
q-1-n=1 \\
q-1+n=55
\end{array}\right.
$$

or
$$
\left\{\begin{array}{l}
q-1-n=5 \\
q-1+n=11
\end{array}\right.
$$

Solving these, we get $\left\{\begin{array}{l}q=29, \\ n=27\end{array}\right.$ or $\left\{\begin{array}{l}q=9 \\ n=3\end{array}\right.$ (discard). Therefore, $q=29, r=31$.