calculating triangle coordinates from 3 lengths and its euler rotation in 3d space I have the following situation:
I have three constant points: $P_1(a_1, b_1, 0), P_2(a_2, b_2, 0), P_3(a_3, b_3, 0)$. These points are on a cartesian 3d space and form an equilateral triangle with a side length equal to len1.
I also have a second equilateral triangle let's call it $T$ (in the same 3d space) with a side length equal to len2 < len1 defined by three points $T_1(x_1, y_1, z_1), T_2(x_2, y_2, z_2), T_3(x_3, y_3, z_3)$ and a constraint: $z_1, z_2, z_3 > 0$. $len2$ is known.
The last known thing are 3 Euler angles: $θ_x, θ_y, θ_z$ which represent the rotation of triangle $Τ$ in the 3d space, $θ_x$ is the rotation of triangle $T$ in the $x$ axis and so on. Constraint: $-90 < θ_x, θ_y, θ_z < 90$ degrees.
The winding order of the triangle is the same, i.e. always $T_1,T_2,T_3$, not possible to have $T_3, T_1, T_2$ or other combinations. The minimum angle of the face normal of triangle $T$ with plane $z=0$ is $>0$ degrees. Just mentioning this to avoid situations like placing the triangle face down.
My first question is:
For all possible values of $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$, i.e. for all possible positions of triangle $T$ with side length =len2 and $z>0$ on a given rotation of the triangle $T$: $θ_x, θ_y, θ_z$, is there a unique vector $(L_1, L_2, L_3)$ where $L_1 = Dist(P_1, T_1), L_2 = Dist(P_2, T_2), L_3 = Dist(P_3, T_3)$ ?
My second question is:
If the answer to the previous question is yes, then given $L_1, L_2, L_3$ and the rotation $θ_x, θ_y, θ_z$ of triangle $Τ$, what are the equations for finding the coordinates of the three points $T_1, T_2, T_3$ of triangle $T$?
Thank you in advance.
 A: Any transformation is decomposable into translation, deformation (here only isotropic deformation is present, due to equilaterality) and rotation. Rotation is done by three subsequent rotations with respect to 3 different axes, where the transformation order matters. The triangle after transformation is given by formula:
\begin{align}
\left[\begin{array}{c}x_i\\y_i\\z_i\end{array}\right]=\left[\begin{array}{c}X_0\\Y_0\\Z_0\end{array}\right]+
\underbrace{\frac{\text{len}_2}{\text{len}_1}
\left[\begin{array}{ccc} 
c_3  & s_3 & 0 \\
-s_3 & c_3 & 0 \\
0    & 0   & 1 \\
\end{array}\right]
\left[\begin{array}{ccc} 
c_2  & 0 & s_2 \\
0    & 1 & 0   \\
-s_2 & 0 & c_2 \\
\end{array}\right]
\left[\begin{array}{ccc} 
1    & 0 & 0 \\
0 & c_1  & s_1   \\
0 & -s_1 & c_1   \\
\end{array}\right]}_{\mathbf{A}}
\left[\begin{array}{c}a_i\\b_i\\0\end{array}\right],~~i=1,2,3,
\end{align}
where $\text{len}_2<\text{len}_1,~z_i>0,~c_i=\cos(\theta_i),~s_i=\sin(\theta_i),~i=1,2,3$, $~\theta_1=\theta_x,~\theta_2=\theta_y,~\theta_3=\theta_z$, where matrix $\mathbf{A}$ is known and where the vector $[X_0,~Y_0,~Z_0]^{\text{T}}$ represents translation.
Continuation:
Moreover we have known distances $L_1,L_2,L_3$ so that we want
\begin{align}
&(x_i-a_i)^2+(y_i-b_i)^2+z_i^2=L_i^2,~~i=1,2,3,
\end{align}
and
\begin{align}
&z_i>0,~~i=1,2,3.
\end{align}
Also we have 12 equations
\begin{align}
&\left[\begin{array}{c}x_i\\y_i\\z_i\end{array}\right]=\left[\begin{array}{c}X_0\\Y_0\\Z_0\end{array}\right]+
\mathbf{A}\left[\begin{array}{c}a_i\\b_i\\0\end{array}\right],~~i=1,2,3,~~~~~~~~~~~~~~~(1)\\
&(x_i-a_i)^2+(y_i-b_i)^2+z_i^2=L_i^2,~~i=1,2,3,~~~~~~(2)\\
\end{align}
for 12 unknowns $X_0,Y_0,Z_0,$ $x_i,y_i,z_i,i=1,2,3$. So we can hope, at least in some cases, in a unique solution. If this solution satisfies $z_i>0,~i=1,2,3$ then we are done. The following is however clear:

*

*There are such input parameters, that problem (1,2) has no solution

*In case $\text{len}_2=\text{len}_1$ there are such input parameters, that problem (1,2) has infinitely many solutions

Proof: 1. for $\mathbf{A}=\mathbf{0},~L_1=L_2=L_3=0$ and different points $P_1,~P_2,~P_3$ the problem definitely has no solution. 2. for $\mathbf{A}=\mathbf{I},~L_1=L_2=L_3=1$ we see that for any $X_0,Y_0,Z_0$ s.t. $X_0^2+Y_0^2+Z_0^2=1$ we get solution, which satifies Eq.(2), ergo we have infinitely many different nontrivial solutions in this case. $\clubsuit$
Remark (uniqueness in case len2<len1): If we insert Eqs.(1) into Eqs.(2) we can solve instead of 12 equations only 3 nonlinear equations:
\begin{align}
&(X_0+(a_{11}-1)a_i+a_{12}b_i)^2+(Y_0+a_{21}a_i+(a_{22}-1)b_i)^2+\\&(Z_0+a_{31}a_i+a_{32}b_i)^2=L_i^2,~~i=1,2,3.~~~~~~(3)\\
\end{align}
This is a problem of intersection of three spheres in $\mathbb{R}^3$, which has in general $0,1,2$ solutions (infinite number of solutions are not possible in case len2<len1). If we can proof, that in case of 2 different solutions one yields negative values of $z_i$ (which is indicated by all numerical solutions, which I have tried), then the proof of uniqueness is made. $\diamondsuit$
