The map $(x, y) \mapsto (100 - x, -y)$ is a $180^\circ$ rotation about $(50, 0)$ that sends the graph of $f$ to the graph of $g$

So the roots pair as $x_2 + x_3 = x_1 + x_4 = 100;$ with $x_3 - x_2 = 150,$ find $x_2, x_3$ and use the vertex condition

Because $g(x) = -f(100 - x),$ the graphs of $f$ and $g$ are reflections of each other through the point $(50, 0),$ so the four intercepts pair up with $x_2 + x_3 = x_1 + x_4 = 100.$

With $x_3 - x_2 = 150,$ we get $x_2 = -25$ and $x_3 = 125.$

Take $x_1, x_3$ as the roots of $f,$ whose vertex has $x$-coordinate $h = \dfrac{x_1 + x_3}{2},$ so $x_1 = 2h - 125.$ The condition that the vertex of $f$ lies on the graph of $g$ gives $$\begin{aligned} 1 &= \frac{f(h)}{g(h)} \\ &= \frac{(125 - h)(h - 125)}{-(h + 25)(3h - 225)}, \end{aligned}$$ which gives $h = -25 \pm 75\sqrt{2}.$ Since $x_1 = 2h-125$ must be less than $x_2=-25,$ we need $h \lt 50,$ so $h = -25-75\sqrt{2}.$

Then $x_4 = 100 - x_1,$ so $$\begin{aligned} x_4 - x_1 &= 350 - 4h \\ &= 450 + 300\sqrt{2}. \end{aligned}$$ Hence $m + n + p = 450 + 300 + 2$ $= 752.$

Thus, the correct answer is D.