jmc

Let $x=\sqrt{10a+14b+22c-d}.$ Then $x^2=10a+14b+22c-d,$ so $d=10a+14b+22c-x^2.$ Then we can write the given equation as $$a^2+b^2+c^2+519=10a+14b+22c-x^2+36x.$$Hence, $$a^2+b^2+c^2+x^2-10a-14b-22c-36x+519=0.$$Completing the square in $a,$ $b,$ $c,$ and $x,$ we get $$(a-5{)}^2+(b-7{)}^2+(c-11{)}^2+(x-18{)}^2=0.$$Therefore, $a=5,$ $b=7,$ $c=11,$ and $x=18.$ Then $$d=10a+14b+22c-x^2=66,$$so $a+b+c+d=5+7+11+66=\boxed{89}.$