Following the usual order of rows, I label the vertices A, B, C, D, E, F and G.
The vertex A has 4 adjacent vertices (B, C, D and E). Other vertices have degree 3 (3 adjacent vertices).
The graph has 7 vertices and 11 edges.
There are the 3-cycles ("triangles"), ACE and ABD.
There are the 4-cycles ("quadrangles") ABGC, ADFE, BDFG and CEFG.
The two admissibility tests for candidate unit distance graphs in 2D are:
(a) Has no subgraph isomorphic to the complete graph on 4 vertices.
This is because in a unit distance graph in the Euclidean plane, we can have 3 points at unit distance from each other, but not 4.
(b) If X and Y are distinct vertices of the graph, there can be two distinct vertices v1 and v2, each adjacent to X and Y, but there cannot be three distinct vertices v1, v2 and v3 each adjacent to X and Y.
Because in the plane, if C1 and C2 are radius 1 circles about two distinct points X and Y, then C1 and C2 intersect in two or fewer points.
I won't check that the 7-graph with the adjacency matrix M has properties (a) and (b), as the program did it, and it's routine.
The graph associated to M is connected also.
Numerically, I found 7 points in R^3 with unit distances to within precision (say 10^(-12) ), for the 11 edges required by the matrix M.
I don't think the graph for M can be for a planar Euclidean figure.
Also, I doubt that the presumed 3-D figure is deformable, I rather think it should be rigid.
If we view ABGC as some sort of base quadrangle, even though it may not fit in a plane, then
upon side AB rises the equilateral triangle ABD to a high point at D; upon side AC rises the equilateral triangle ACE to a high point at E; connected to the quadrangular base ABGC of the skeleton figure along side BG is the quadrangle BDFG; also connected to base ABGC along side CG is the quadrangle CEFG. The quadrangle ADFE contains the base point A and the three so-called high-points D, E and F, all in quadrangle ADFE but not in the base quadrangle ABGC.
The presumed 3-D figure may have a plane of symmetry passing through the three points A, F and G.
This plane of symmetry would send B to C and C to B, and send D to E and E to D. It would also keep fixed the three points A, F and G. There might be no plane of symmetry if the presumed 3D figure is deformable; I doubt that it's deformable.
From my sketch of the graph, I think it's clear that there's a graph isomorphism that fixes A, G and F and that permutes B with C, and D with E.
It's only after much numerical computer work that I have some confidence that there are 7 points in R^3 which are at distance 1 when implied by matrix M; if a matrix entry in M is zero, the corresponding distance might or might not be 1: it doesn't matter as I view the problem.
So, I'd be curious about ways to show the corresponding graph to M can't be made a 2D unit distance graph, can be made a 3D equivalent of a unit distance graph, and so on.
Here are the x, y and z coordinates for points A to G in the presumed 3-D figure:
x y z 1.384667538045070848 1.382768373981428755 0.044412611314774557 // A 0.399833591130468233 1.520541863804588859 0.149866689028978084 // B 0.970226156884894247 1.649162807406999439 0.914626577774698943 // C 1.041203194674366289 2.053898061776666813 0.701388400056488421 // D 0.405996262469059331 1.582686954821862589 0.091689406861527141 // E 1.301674189203176470 1.182510983474254957 0.285648793118733590 // F 1.172171851884524849 2.151166533083443602 0.073668674144507970 // G
P.S. If we have a square base ABGC in R^2, and one unit above it a square A'DFE, and add in the 12 edges, then remove the edge AA' and glue A and A' while letting the other edges stretch, we're left with a 7-vertex, 11-edge graph that looks like the graph of matrix M.