Constraints on 3D point locations admitting $n$ planes of a given distance from each This question is an extension of this prior one, focussing on the conditions under which a solution to that prior problem is possible.  That prior problem considered the case of three points in three dimensions, each far from each other and not approximately colinear, and the number of planes that were a unit distance from all three points.  In brief, the answer is $2^3 = 8$, corresponding to planes tangent to the "top" of all three unit spheres centered on the points; on the "bottom" of all three unit spheres; on the "top" of sphere $A$ and the "bottom" of $B$ and $C$, and so forth.  (A figure in one solution of the problem may help you visualize what is going on.)
The current question centers on the constraints on the locations of the three points in three dimensions that admit this solution of eight planes each of which is a unit distance from the three points.  Clearly, if the points are too close together (e.g., distance less than 2), the above solution cannot hold, as one cannot find planes that are tangent to all configurations of "top" and "bottom" for the nearby spheres.  But what are the full sets of conditions that admit the eight planes?
Without loss of generality, assume the three points in three dimensions are constrained to the $(x,y)$ plane:  point $A$ at the origin $(0,0,0)$, point $B$ on the $x$ axis $(x_B,0,0)$, and point $C$ in the upper half plane $(x_C,y_C,0)$ where $y_C \ge 0$.  
What relations among the variables admit all eight planes?  What relations among the variables admit exactly four planes?  (Note that if the points are co-linear--i.e., $y_C=0$--then there is an infinite number of planes.)
Addendum
@Rahul states that sphere $A$ must lie outside the cones defined by spheres $B$ and $C$ for all eight tangent planes to exist.  This is not sufficient, as this figure illustrates:
 
The blue sphere lies outside the cone(s) defined by the green and red, yet one cannot obtain all eight tangent planes.
 A: In my original answer I overcomplicated the problem by generalizing it to spheres of varying radii. For the case of unit radii the solution is much simpler. As in the question, we take the three points $A,B,C$ to lie in the $xy$-plane, with the $x$-axis along $AB$. Consider the three altitudes of the triangle $ABC$.
$$\begin{align}
\text{number of tangent planes} &= 2 \\
&+ 2 \times (\text{number of altitudes of length > 2}) \\
&+ (\text{number of altitudes of length 2})
\end{align}$$
The first term counts the two planes $z=\pm 1$; these are always tangent to the three spheres.
For the second and third terms, consider the altitude from $C$, whose length is $|y_C|$. In the $yz$-plane, the spheres around $A$ and $B$ both project to the unit circle centered at $(0,0)$, and the sphere around $C$ projects to a unit circle centered at $(y_C,0)$. The two circles are disjoint, tangent, or intersecting depending on whether $|y_C|>2$, $|y_C|=2$, or $|y_C|<2$, in which case they have two, one, or zero internal tangents respectively (excluding the external tangents $z=\pm 1$). These correspond to two, one, or zero planes tangent to all three spheres.
In particular, if the three spheres are very close to each other, they form a triangle whose side lengths are nearly $2$, and there are no altitudes of length $\ge 2$. Thus we correctly predict only $2$ tangent planes, namely the ones entirely above or below the spheres.
