The distance between the parallel lines given by ${(x+7y)}^2+4\sqrt{2}(x+7y)-42=0$ is

$x^2+14xy+49y^2+4\sqrt{2}x+28\sqrt{2}y-42=0$

$a=1,b=49,2h=14,2g=4\sqrt{2}$

$2f=28\sqrt{2},c=-4;2\sqrt{\frac{g^2-ac}{a(a+c)}}=2$