smc

Note that $f(x)-x^2$ has roots $2,3$, and $4$. Therefore, we may write $f(x)=a(x-2)(x-3)(x-4)+x^2$. Now we find that lines $AB$, $AC$, and $BC$ are defined by the equations $y=5x-6$, $y=6x-8$, and $y=7x-12$ respectively. Since we want to find the $x$-coordinates of the intersections of these lines and $f(x)$, we set each of them to $f(x)$ and synthetically divide by the solutions we already know exist. In the case of line $AB$, we may write $a(x-2)(x-3)(x-4)+x^2-5x+6=a(x-2)(x-3)(x-r_1)$ for some real number $r_1$. Dividing both sides by $(x-2)(x-3)$ gives $a(x-4)+1=a(x-r_1)$ or $r_1=\frac{4a-1}{a}$. For line $AC$, we have $a(x-2)(x-3)(x-4)+x^2-6x+8=a(x-2)(x-4)(x-r_2)$ for some real number $r_2$, which gives $a(x-3)+1=a(x-r_2)$ or $r_2=\frac{3a-1}{a}$. For line $BC$, we have $a(x-2)(x-3)(x-4)+x^2-7x+12=a(x-3)(x-4)(x-r_3)$ for some real number $r_3$, which gives $a(x-2)+1=a(x-r_3)$ or $r_3=\frac{2a-1}{a}$. Since $r_1+r_2+r_3=24$, we have $\frac{4a-1}{a}+\frac{3a-1}{a}+\frac{2a-1}{a}=24$ or $\frac{9a-3}{a}=24$. Solving for $a$ gives $a=-\frac{1}{5}$. Substituting this back into the original equation, we get $f(x)=-\frac{1}{5}(x-2)(x-3)(x-4)+x^2$, and $f(0)=-\frac{1}{5}(-2)(-3)(-4)+0=\boxed{\frac{24}{5}}$