WLOG $p\geq q$ If $p,q,r>7$: $r(r-7)>p(p-7)\Leftrightarrow r>p$. $p(p-7)=r^2-7r-q^2+7q=(r-q)(r+q-7)\Rightarrow p|(r-q)(r+q-7)$ If $p\mid r-q$ then $r-q\geq p$ so $r+q-7\leq p-7<p\leq r-q\Rightarrow 2q<7$,contradiction. So $p|r+q-7\Rightarrow r+q=kp+7,k\in\mathbb{N}\Leftrightarrow r=kp+7-q\geq (k-1)p+7$ Also $r(r-7)=p(p-7)+q(q-7)\leq 2p(p-7)$.So if $k\geq 3$ then $r\geq 2p+7$ and $(2p+7)2p\leq p(p-7)\Leftrightarrow 4p+14\leq p-7$,contradiction. Therefore $k=0,1,2$ $k=0\Rightarrow r+q=7$,contradiction $k=1\Rightarrow r+q=p+7<r+7\Rightarrow q<7$,contradiction $k=2\Rightarrow r+q=2p+7$.So $p(p-7)=(r-2p-7+r)2p\Leftrightarrow 5p+7=4r\Rightarrow $ $\Rightarrow -p+1\equiv r\pmod3 \Rightarrow p,r\equiv 2\pmod 3$ Therefore $q=2p+7-r\equiv 4+7-2\equiv0\pmod3$,contradiction. So we must have $r=2,3,5,7$. $p^2-7p-r^2+7r=-q(q-7)\Leftrightarrow (p-r)(p+r-7)=q(7-q)$ $q=7\Rightarrow p=r\,\,or \,\,\,p+r=7\Leftrightarrow \boxed{(p,q,r)=(a,7,a),(5,2,7)\,\,where\,\,a\,\,is \,\,prime}$ $q=5\Rightarrow (p-r)(p+r-7)=10\Leftrightarrow \boxed{(p,q,r)=(7,5,5),(7,5,2)}$ $q=3\Rightarrow (p-r)(p+r-7)=21\Leftrightarrow \boxed{(p,q,r)=(7,3,3)}$ $q=2\Rightarrow (p-r)(p+r-7)=10\Leftrightarrow \boxed{(p,q,r)=(7,5,5),(7,5,2)}$ So the solutions are $\boxed{(p,q,r)=(2,5,7),(2,5,7),(7,5,5),(5,7,5),(5,7,2),(7,5,2),(7,3,3),(3,7,3),(a,7,a),(7,a,a)\,\,where\,\,a\,\,is \,\,prime}$ I hope I do not forget any solution