Answer: 306

Small Hint:

Normalize the perpendicular distances from $P$ to the three sides

Big Hint:

A cross-section parallel to a side has length equal to that side times one minus the corresponding normalized distance

Solution:

Write $a=BC=450,$ $b=CA=510,$ and $c=AB=425.$ Let $x,$ $y,$ and $z$ be the distances from $P$ to $BC,$ $CA,$ and $AB,$ respectively, each divided by the corresponding altitude. Area decomposition gives $x+y+z=1.$

By similar triangles, the segment through $P$ parallel to $BC$ has length $a(1-x),$ and similarly the other two lengths are $b(1-y)$ and $c(1-z).$ Since all three equal $d,$ $$\begin{aligned} x&=1-\frac da,\\ y&=1-\frac db,\\ z&=1-\frac dc. \end{aligned}$$ Their sum is $1,$ so $$d=\frac{2abc}{ab+bc+ca}.$$ Substituting the three side lengths gives $$\begin{aligned} d&=\frac{195075000}{637500}\\ &=306. \end{aligned}$$