The given equality rewrites as $p^2 + q^2 - r^2 = 7(p + q - r)$. Since $(p+q-r)(p+q-r) = p^2 + q^2 - r^2 + 2pq$, it follows that $p+q-r$ divides $2pq$.
If $p$, $q$, $r > 2$, then $p+q-r$ is odd, so $p+q-r = p$, $q$ or $pq$. The first case gives $r = q$, then $p = 7$, so $(p, q, r) = (7, q, q)$, $q$ being an arbitrary prime. Likewise, the second case gives $(p, q, r) = (p, 7, p)$, $p$ being an arbitrary prime. If $p+q-r = pq$, then $1-r = (p-1)(q-1)$, impossible.
If $p=2$ (or $q=2$) the equality becomes $(q-r)(q+r-7) = 10$, implying $q=7$ and $r=5$, with two new solutions obtained, $(p, q, r) = (2, 7, 5)$ and $(p, q, r) = (7, 2, 5)$, or $r=2$, with no further solutions obtained.
The solution triplets therefore are $(2, 7, 5)$, $(7, 2, 5)$, $(7, t, t)$, $(t, 7, t)$, where $t$ is an arbitrary prime. (Notice the fact $r$ was known to be a prime turned out to be inconsequential).