This is the graph of the formula for an area of 25 solved for one of the variables. I noticed for most of the solution area the graph is pretty flat. So I approximated it with this:
I did that by putting a plane through the points:
These are valid points on the plane I'm looking for no matter the area A. That took a while to figure out.
Finding the plane that goes through those 3 points and simplifying yields:
It is linear in x,y,z for a constant A. That should make a few things more easily possible than with the original Heron's formula without losing much accuracy in the domain of possible triangles.