Finding tensegrity forms...(MdG)

The thick lines AB, CD, EF, are rigid bars that are joined in space (and somehow paradoxically kept apart, no one touching another) by means of 9 flexible inextensible cables (thin lines in the figure) joining the bars vertices. The cables are stretched in such a way that the whole structure is rigid. The length of AC, CE, EA, BD, DF, FB is the same and also the other three cables AF, CB, ED have a common length. The triangles ACE and BDF are equilateral and located in parallel planes. The line joining their centers is perpendicular to these planes so that the triangle BDF is obtained by lowering the triangle ACE and rotating it 30¼. There is another similar structure which is a mirror image of this one.

Problem 1. (A problem related to the generalization of the oblique triangular prism)
We are given five concrete points in space A (0,0,0), B (1,1,1), C (0,1,0), D (1,0,0), E (0,0,1) and consider another sixth point F(a,b,c), to be determined under the following conditions. Consider the twelve segments AB, CD, EF, AD, DE, EA, CF, FB, BC, AC, BE, DF (observe that this is the topological structure of segments in the oblique triangular prism).
Find whether it is possible to situate point F in such a way that we get a rigid structure after substituting some of the segments by bars or cables under adequate tensions.

Theorem 3 . Let us consider the spatial points 1,2,3,4,5,6 and assume that no four of them are coplanar.

Let us consider the configuration of points and segments that has the given points as nodes and the segments 12,23,31,45,56,61,14,15,25,26,36,34 (i.e. the essentially (combinatorially) unique configuration with these 6 points and 4 segments concurring at each point).

Consider the (unique) ruled hyperboloid passing through 1,2,4,5,6, whose tangent planes at points 1,2,4 and 5 are 124, 245, 451 and 512, respectively.

Then the proposed configuration admits a tensegrity structure if and only if point 3 lies on this hyperboloid.