2017 AMC 12B Problem 23

Since $A, B, C$ lie on $y = x^2,$ let $g(x) = f(x) - x^2,$ so $g(x) = a(x-2)(x-3)(x-4)$

A line meets the cubic where $f(x) - L(x) = 0;$ by Vieta the three $x$-coordinates of each such triple sum to $9 - \tfrac1a$

The points $A, B, C$ lie on $y = x^2,$ so $g(x) = f(x) - x^2$ has roots $2, 3, 4:$ $g(x) = a(x-2)(x-3)(x-4)$ for some $a \ne 0.$ The coefficients of $x^3$ and $x^2$ in $f$ are $a$ and $1 - 9a,$ so by Vieta the three roots of $f(x) - L(x)$ (for any linear $L$) sum to $9 - \tfrac1a.$ The lines $AB, AC, BC$ meet the cubic in triples $\{2, 3, x_D\},$ $\{2, 4, x_E\},$ $\{3, 4, x_F\},$ so $$\begin{aligned} &x_D + x_E + x_F \\ &\quad {}= 3\left(9 - \tfrac1a\right) \\ &\quad {}- 2(2 + 3 + 4) \\ &\quad {}= 9 - \tfrac3a = 24, \end{aligned}$$ giving $a = -\tfrac15.$ Then $f(x) = x^2$ $- \tfrac15(x-2)(x-3)(x-4),$ so $f(0) = 0 - \tfrac15(-2)(-3)(-4)$ $= \tfrac{24}{5}.$

Thus, the correct answer is D.