Recall the focus and the directrix are such that the parabola is the locus of points equidistant from the focus and the directrix. We will consider pairs of parabolas and find their points of intersections (we label counterclockwise): (1): $P_{1} \cap P_{2}$, two parabolas with directrices adjacent edges on the hexagon (sharing vertex $A$ ). The intersection inside the hexagon can be found by using similar triangles: by symmetry this $X$ must lie on $O A$ and must have that its distance from $A B$ and $F A$ are equal to $|O X|=x$, which is to say $$\sin 60^{\circ}=\frac{\sqrt{3}}{2}=\frac{x}{|O A|-x}=\frac{x}{1-x} \Longrightarrow x=2 \sqrt{3}-3$$ By symmetry also, the second intersection point, outside the hexagon, must lie on $O D$. Furthermore, $X$ must have that its distance $A B$ and $F A$ are equal to $|O X|$. Then again by similar triangles $$\sin 60^{\circ}=\frac{\sqrt{3}}{2}=\frac{x}{|O A|+x}=\frac{x}{1+x} \Longrightarrow x=2 \sqrt{3}+3$$ (2): $P_{1} \cap P_{3}$, two parabolas with directrices edges one apart on the hexagon, say $A B$ and $C D$. The intersection inside the hexagon is clearly immediately the circumcenter of triangle $B O C$ (equidistance condition), which gives $$x=\frac{\sqrt{3}}{3}$$ Again by symmetry the $X$ outside the hexagon must lie on the lie through $O$ and the midpoint of $E F$; then one can either observe immediately that $x=\sqrt{3}$ or set up $$\sin 30^{\circ}=\frac{1}{2}=\frac{x}{x+\sqrt{3}} \Longrightarrow x=\sqrt{3}$$ where we notice $\sqrt{3}$ is the distance from $O$ to the intersection of $A B$ with the line through $O$ and the midpoint of $B C$. (3): $P_{1} \cap P_{4}$, two parabolas with directrices edges opposite on the hexagon, say $A B$ and $D E$. Clearly the two intersection points are both inside the hexagon and must lie on $C F$, which gives $$x=\frac{\sqrt{3}}{2}$$ These together give that the sum desired is $$6(2 \sqrt{3}-3)+6(2 \sqrt{3}+3)+6\left(\frac{\sqrt{3}}{3}\right)+6(\sqrt{3})+6\left(\frac{\sqrt{3}}{2}\right)=35 \sqrt{3}$$