The real numbers $a,$ $b,$ $c,$ and $d$ satisfy
\[a^2 + b^2 + c^2 + 519 = d + 36 \sqrt{10a + 14b + 22c - d}.\]Find $a + b + c + d.$
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 $a + b + c + d = 5 + 7 + 11 + 66 = \boxed{89}.$