They'd have to build 4, actually. Imagine 3 intersecting spheres, there would be 2 possible positions remaining (just as with 2d triangulation, having 2 circles

They'd have to build 4, actually. Imagine 3 intersecting spheres, there would be 2 possible positions remaining (just as with 2d triangulation, having 2 circles would leave 2 points remaining and we'd need a 3rd to find out which was correct