jmc

The intersection of the asymptotes is $(1,2),$ so this is the center of the hyperbola. Since the slopes of the asymptotes are $\pm 1,$ the equation of the hyperbola can be written in the form $$(x-1{)}^2-(y-2{)}^2=d$$for some constant $d.$ Setting $x=3$ and $y=3,$ we get $d=3,$ so the equation is $$\frac{(x-1{)}^2}{3}-\frac{(y-2{)}^2}{3}=1.$$Then $a^2=3$ and $b^2=3,$ so $c^2=a^2+b^2=6,$ which means $c=\sqrt{6}.$ Therefore, the distance between the foci is $2c=\boxed{2\sqrt{6}}.$