Prove that no plane passing through origin in $\mathbb R^3$ can contain all the six open balls of radius $\frac{1}{30}$ unit, one centred at each of the points $(\pm 1, 0, 0), (0, \pm 1, 0), (0, 0 , \pm 1)$, on the same side of the half space.
It seems very intuitive because a plane divides the whole space in two half spaces. Since all the open balls are centered at the distance of $1$ unit from the origin with radius $\frac {1}{30}$ on the co-ordinate axes, none of them passes through origin or touches the other open ball. So, no matter which plane passing through origin we take, some ball will remain totally in the other half space as compared to the rest of the 5 open balls.
But I am having hard time proving it rigorously. Any help will be appreciated. Thanks.
Let's consider first the problem being restricted to the 6 center points only.
Points (0,0,0) and ($\pm$1,0,0) are on a single line. Thus any plane through the origin, either hits that line, and then we have alraidy a point on either side, or does contain that entire line.
Next we consider the orthogonal line through the points (0,0,0) and (0,$\pm$1,0). The plane, alraidy containing the former line, again could either hit that second line, and then we have points on either side, or could contain that second entire line. But because those 2 lines are orthogonal, the plane now has got fixed, i.e. has no more degrees of freedom.
Therefore the third orthogonal line through (0,0,0) and (0,0,$\pm$1) surely would provide points on either side.
So far this argument does consider just the 6 points, not the tiny balls. But because those balls would contain their center points as well, the above impossibility would discard your problem with the much larger set of points as well.
--- rk