jmc

We have that $$\frac{x^2-19}{x^3-2x^2-5x+6}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}.$$Multiplying both sides by $x^3-2x^2-5x+6=(x-1)(x+2)(x-3),$ we get $$x^2-19=A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2).$$Setting $x=1,$ we get $-6A=-18$, so $A=3.$

Setting $x=-2,$ we get $15B=-15,$ so $B=-1.$

Setting $x=3,$ we get $10C=-10,$ so $C=-1.$ Hence, $ABC=\boxed{3}.$