I have $n$ points in $\mathbb{R}^3$ with one point called 'central'. I can write out side lengths of all triangles that contains this central point (side that is not lie on central point always comes at the end), e.g. for $n=4$ and (1) as central point we can get: $$(a_{12}, a_{13}, a_{23}), (a_{12}, a_{14}, a_{24}), (a_{13}, a_{14}, a_{34}).$$ Overall, it will be $\frac{(n-1)(n-2)}{2}$ triplets of distances. It is obvious, that if I keep information about indices, I can restore pairwise distances between all points. By knowing pairwise distances, I can get the structural information (coordinates in some orthogonal basis, order of points is not important).
The question is, what if I lost this information (side without central point still at the end), i.e. have only $$(d_{1}, d_{2}, d_{3}), (d_{1}, d_{4}, d_{5}), (d_{2}, d_{4}, d_{6})$$ would it be still possible to restore structure?
For the case of $d_1 \neq \cdots \neq d_{\frac{n(n-1)}{2}}$ it is true, because I can easily assign indices and restore pairwise distances. If some distances are equal, I don't know what to do. Can you give a hint?
I think you should be OK as long as you have at least two unique central distances.
Pick some triangle $t$ with unique central distances $x$ and $y$ and place it on the plane, with the central vertex at an arbitrary origin, at an arbitrary angular orientation and with arbitrary chirality. Since you only want pairwise distances and the locations of the points are only known up to isometries, all these choices don't matter and merely fix your coordinate system.
Now find the other $n-3$ triangles that contain $x$ and the other $n-3$ triangles that contain $y$. For a $k$-fold central distance $z$, there will be $k$ triangles with central distances $x$ and $z$ (“$xz$-triangles”) and $k$ triangles with central distances $y$ and $z$ (“$yz$-triangles”). Each of these can be placed in one of two chirally distinguished ways such that the $x$ or $y$ side coincides with the corresponding side of $t$. Unless there are further coincidences in the positions, only the “correct” chiralities and the “correct” pairings of the $xz$-triangles with the $yz$-triangles will make the points at the end of the sides of length $z$ come out consistently for each pair. This allows you to correctly place all these triangles; then you have all the points and thus also all the pairwise distances.