Use the parametrization $a=k(m^2-n^2), b=k(2mn), c=k(m^2+n^2)$ and $6=\frac{k^2(m^2-n^2)(mn)}{k(m^2+mn)}=kn(m-n)$. Just cases now.

MathLuis wrote: Find all right triangles with integer sides and inradius 6. Let the legs be $a$ and $b$ and the hypotenuse $c$: $\Rightarrow a+b=c+2(6)=c+12$ we also have $a^2+b^2=c^2$ Substituting we get that: $ab=12c+72$ $\Rightarrow (a-12)(b-12)=72$ and cases