Now one can hope that geometric considerations will show that (in this "generic" situation) the intersection class in $H^6((\mathbb{RP}^2)^3,\mathbb{Z}/2\mathbb{Z})$ will be nontrivial, and that furthermore that there is no contribution from some degenerate case e.g. two planes coinciding. Then it probably still remains necessary to pass to some perturbation of $K$, and to run this argument there instead. So even if this sketch can be made to work, I doubt that a fully rigorous proof would fit in a half-page...

Some naive thoughts: We can try and derive this from an intersection theory statement on $(\mathbb{RP}^2)^3$. Points correspond to triples of planes, and the condition that three planes split $K$ into $8$ parts of equal volume will give a codimension $3$ cycle for "generic" $K$. (We get codimension $3$ instead of codimension $7$ because diametrically opposite regions have the same area, by reflection symmetry of $K$). Similarly each of the conditions that two planes split a cross-section into four parts of equal area should lead to a codimension $1$ cycle.

There are notes of Haiman I like for this topic; I think it might be this: math.berkeley.edu/~mhaiman/ftp/newt-sf-2001/newt.pdf. In particular, the case of your question is addressed in section 3.1.

@JasonStarr Does the Milnor-Thom approach provide a bound for any subvariety? I was under the impression that their theorem only covered the cases of hypersurfaces.

Thank you for this reference! I will leave the question open for a bit to see if I get any other answers, but this seems to completely answer my question.

@Libli: The issue is that the Hilbert scheme may become highly disconnected when one removes points corresponding to singular varieties - I admittedly don't have an example of the top of my head, but I would be extremely surprised if one did not exist.

@AndrésE.Caicedo Is it easy to write down what the relevant rational approximations are? I don't see how to do it immediately.

On a somewhat different note, I've always been curious if anyone has drawn pictures of $Bun_G$.

I believe you can prove this just by looking at the map $\widehat{S}\rightarrow C$ and observing that it's generally smooth. It follows that the exceptional curves must lie in finitely many fibers, but each of those fibers is dimension $1$ and finite type.

@BenCrowell: I don't see how - I'm interpreting his answer as showing that for any choice of $a$ and $b$ with $a-b$ irrational, most swaps will lead in transcendentals, while I'm showing that for most choices of $a,b$, all swaps will lead to transcendentals.