Find constants $A,$ $B,$ and $C$ so that
\[\frac{x^2 - 7}{(x - 2)(x - 3)(x - 5)} = \frac{A}{x - 2} + \frac{B}{x - 3} + \frac{C}{x - 5}.\]Enter the ordered triple $(A,B,C).$
Multiplying both sides by $(x - 2)(x - 3)(x - 5),$ we get
\[x^2 - 7 = A(x - 3)(x - 5) + B(x - 2)(x - 5) + C(x - 2)(x - 3).\]Setting $x = 2,$ we get $3A = -3,$ so $A = -1.$

Setting $x = 3,$ we get $-2B = 2,$ so $B = -1.$

Setting $x = 5,$ we get $6C = 18,$ so $C = 3.$  Thus, $(A,B,C) = \boxed{(-1,-1,3)}.$