Is it possible to get an explicitly formula for the following change of variable (formula for the Jacobian or for the inverse. I would even accept results from mathematica or other software, which I never use and I'm not proficient in).
Let me first introduce the following function $\omega : \left\lbrace \begin{aligned} \mathbb{R}^3 & \longrightarrow \ \mathbb{R}\\ \mathbf{k}\; & \longmapsto \sqrt{\mathbf{k}^2 + m^2} \end{aligned} \right. $. I'm now considering $$ \int_{\mathbb{R}^{3\times 3}} \frac{ \varphi\big(\omega(\mathbf{k}_1) + \omega(\mathbf{k}_2) +\omega(\mathbf{k}_3), - \mathbf{k}_1 - \mathbf{k}_2 - \mathbf{k}_3 \big)}{ \omega(\mathbf{k}_1)\hspace{1pt} \omega(\mathbf{k}_2) \hspace{1pt} \omega(\mathbf{k}_3)\, \big[ \big(\omega(\mathbf{k}_1) + \omega(\mathbf{k}_2) + \omega(\mathbf{k}_3) \big)^2 - \omega(\mathbf{k}_1+ \mathbf{k}_2 + \mathbf{k}_3)^2 \big]^2}\, d\mathbf{k}_1\,d\mathbf{k}_2\, d\mathbf{k}_3 $$ where $\varphi$ is a function of 4 variables. It seems natural to consider
$$ \Phi: \left\lbrace \begin{aligned} U \subset \mathbb{R}^9 & \longrightarrow V \subset \mathbb{R}^3 \times [m,+\infty[^3 \times ]0,\pi[^3 \\ \begin{pmatrix} \mathbf{k}_1 \\ \mathbf{k}_2 \\ \mathbf{k}_3 \end{pmatrix}\ & \longmapsto \begin{pmatrix} \mathbf{k} \\ \omega_{\mathbf{k}_1} \\ \omega_{\mathbf{k}_2} \\ \omega_{\mathbf{k}_3}\\ \theta_{12} \\ \theta_{23} \\ \theta_{31} \end{pmatrix} := \begin{pmatrix} \mathbf{k}_1 + \mathbf{k}_2 + \mathbf{k}_3\\ \omega(\mathbf{k}_1) \\ \omega(\mathbf{k}_2) \\ \omega(\mathbf{k}_3)\\ \arccos \big( \mathbf{k}_1 \cdot \mathbf{k}_2 \big/ \left\lVert\mathbf{k}_1 \right\rVert \left\lVert\mathbf{k}_2\right\rVert \big)\\ \arccos\big( \mathbf{k}_2 \cdot \mathbf{k}_3 \big/ \left\lVert\mathbf{k}_2\right\rVert\left\lVert\mathbf{k}_3\right\rVert \big)\\ \arccos\big( \mathbf{k}_3 \cdot \mathbf{k}_1 \big/ \left\lVert\mathbf{k}_3\right\rVert \left\lVert\mathbf{k}_1\right\rVert \big) \end{pmatrix} \end{aligned} \right. $$ where $U:= \left\lbrace (\mathbf{k}_1 , \mathbf{k}_2 , \mathbf{k}_3)\in \mathbb{R}^9,\ \operatorname{det}(\mathbf{k}_1 , \mathbf{k}_2 , \mathbf{k}_3) >0 \right\rbrace$ is just half the space and the image $V$ a little complicated.
I'm not entirely sure this a $\mathcal{C}^1$-diffeomorphism. The Jacobian looks like $$ \begin{vmatrix} 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\ k_1^1 / \omega_{\mathbf{k}_1} & k_1^2 / \omega_{\mathbf{k}_1} & k_1^3 / \omega_{\mathbf{k}_1} & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & k_2^1 / \omega_{\mathbf{k}_2} & k_2^2 / \omega_{\mathbf{k}_2} & k_2^3 / \omega_{\mathbf{k}_2} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & k_3^1 / \omega_{\mathbf{k}_3} & k_3^2 / \omega_{\mathbf{k}_3} & k_3^3 / \omega_{\mathbf{k}_3} \\ \frac{\cos \theta_{12}\hspace{.8pt} \frac{k_1^1}{\lVert\mathbf{k}_1\rVert}- \frac{k_2^1}{\lVert\mathbf{k}_2\rVert}}{\lVert\mathbf{k}_1\rVert\, \sin \theta_{12}} & '' %\frac{\cos \theta_{12}\, \frac{k_1^2}{\norm{\mathbf{k}_1}}- \frac{k_2^2}{\norm{\mathbf{k}_2}}}{\norm{\mathbf{k}_1} \sin \theta_{12}} & '' % \frac{\cos \theta_{12}\, \frac{k_1^3}{\norm{\mathbf{k}_1}}- \frac{k_2^3}{\norm{\mathbf{k}_2}}}{\norm{\mathbf{k}_1} \sin \theta_{12}} & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{\cos \theta_{23}\hspace{.8pt} \frac{k_2^1}{\lVert\mathbf{k}_2\rVert}- \frac{k_3^1}{\lVert\mathbf{k}_3\rVert}}{\lVert\mathbf{k}_2\rVert\, \sin \theta_{23}} & '' & '' & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{\cos \theta_{31}\hspace{.8pt} \frac{k_3^1}{\lVert\mathbf{k}_3\rVert}- \frac{k_1^1}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_3\rVert\, \sin \theta_{31}} & '' & '' \end{vmatrix}$$ where $''$ stands for similar and not identical and where I used the convention from physics that $k^i_j$ is the $i^{\text{th}}$-component of $\mathbf{k}_j$ and not something to the power $i$.
Finally I get $$\operatorname{det}\,\left( \mathrm{d}\Phi \right) = \frac{\operatorname{det}^2\big(\mathbf{k}_1 , \mathbf{k}_2 , \mathbf{k}_3\big)}{\omega_{\mathbf{k}_1}\,\omega_{\mathbf{k}_2}\,\omega_{\mathbf{k}_3}\, \lVert\mathbf{k}_1\rVert^2\hspace{.9pt}\lVert\mathbf{k}_2\rVert^2 \hspace{.9pt}\lVert\mathbf{k}_3\rVert^2\sin \theta_{12}\, \sin \theta_{23}\, \sin \theta_{31}}$$ and I suspect the determinant is exactly the product norm of $\mathbf{k}_i \times \sin \theta_{ij}$. Here are the details of my calculations. One could in fact simplify further before introducing the $\Delta_{99}$.
1) Determinant: let us denote it $J_{\Phi}$. By line operations, we obtain $$ J_{\Phi} = \begin{vmatrix} 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & - k_1^1 / \omega_{\mathbf{k}_1} & - k_1^2 / \omega_{\mathbf{k}_1} & - k_1^3 / \omega_{\mathbf{k}_1} & - k_1^1 / \omega_{\mathbf{k}_1} & - k_1^2 / \omega_{\mathbf{k}_1} & - k_1^3 / \omega_{\mathbf{k}_1}\\ 0 & 0 & 0 & k_2^1 / \omega_{\mathbf{k}_2} & k_2^2 / \omega_{\mathbf{k}_2} & k_2^3 / \omega_{\mathbf{k}_2} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & k_3^1 / \omega_{\mathbf{k}_3} & k_3^2 / \omega_{\mathbf{k}_3} & k_3^3 / \omega_{\mathbf{k}_3} \\ 0 & 0 & 0 & \frac{\frac{k_2^1}{\lVert\mathbf{k}_2\rVert} - \cos \theta_{12}\hspace{.8pt} \frac{k_1^1}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_1\rVert\, \sin \theta_{12}} & '' & '' & \frac{\frac{k_2^1}{\lVert\mathbf{k}_2\rVert} - \cos \theta_{12}\hspace{.8pt} \frac{k_1^1}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_1\rVert\, \sin \theta_{12}} & '' & ''\\ 0 & 0 & 0 & \frac{\cos \theta_{23}\hspace{.8pt} \frac{k_2^1}{\lVert\mathbf{k}_2\rVert}- \frac{k_3^1}{\lVert\mathbf{k}_3\rVert}}{\lVert\mathbf{k}_2\rVert\, \sin \theta_{23}} & '' & '' & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{\cos \theta_{31}\hspace{.8pt} \frac{k_3^1}{\lVert\mathbf{k}_3\rVert}- \frac{k_1^1}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_3\rVert\, \sin \theta_{31}} & '' & '' \end{vmatrix} \\ $$
Develop w.r.t the last line, denoting $\Delta_{ij}$ appropriate minor $$J_{\Phi} = \frac{\cos \theta_{31}\hspace{.8pt} \frac{k_3^3}{\lVert\mathbf{k}_3\rVert}- \frac{k_1^3}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_3\rVert\, \sin \theta_{31}}\, \Delta_{99} - \frac{\cos \theta_{31}\hspace{.8pt} \frac{k_3^2}{\lVert\mathbf{k}_3\rVert}- \frac{k_1^2}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_3\rVert\, \sin \theta_{31}}\, \Delta_{98} + \frac{\cos \theta_{31}\hspace{.8pt} \frac{k_3^1}{\lVert\mathbf{k}_3\rVert}- \frac{k_1^1}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_3\rVert\, \sin \theta_{31}}\, \Delta_{97}$$
Now $$ \Delta_{99} = \begin{vmatrix} - k_1^1 / \omega_{\mathbf{k}_1} & - k_1^2 / \omega_{\mathbf{k}_1} & - k_1^3 / \omega_{\mathbf{k}_1} & - k_1^1 / \omega_{\mathbf{k}_1} & - k_1^2 / \omega_{\mathbf{k}_1} \\ k_2^1 / \omega_{\mathbf{k}_2} & k_2^2 / \omega_{\mathbf{k}_2} & k_2^3 / \omega_{\mathbf{k}_2} & 0 & 0 \\ 0 & 0 & 0 & k_3^1 / \omega_{\mathbf{k}_3} & k_3^2 / \omega_{\mathbf{k}_3} \\ \frac{\frac{k_2^1}{\lVert\mathbf{k}_2\rVert} - \frac{\cos \theta_{12}\hspace{.8pt} k_1^1}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_1\rVert\, \sin \theta_{12}} & \frac{\frac{k_2^2}{\lVert\mathbf{k}_2\rVert} - \frac{\cos \theta_{12}\hspace{.8pt} k_1^2}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_1\rVert\, \sin \theta_{12}} & \frac{\frac{k_2^3}{\lVert\mathbf{k}_2\rVert} - \frac{\cos \theta_{12}\hspace{.8pt} k_1^3}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_1\rVert\, \sin \theta_{12}} & \frac{\frac{k_2^1}{\lVert\mathbf{k}_2\rVert} - \frac{ \cos \theta_{12}\hspace{.8pt} k_1^1}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_1\rVert\, \sin \theta_{12}} & \frac{\frac{k_2^2}{\lVert\mathbf{k}_2\rVert} - \frac{\cos \theta_{12}\hspace{.8pt} k_1^2}{\lVert\mathbf{k}_1\rVert}}{\lVert\mathbf{k}_1\rVert\, \sin \theta_{12}} \\ \frac{\frac{\cos \theta_{23}\hspace{.8pt} k_2^1}{\lVert\mathbf{k}_2\rVert}- \frac{k_3^1}{\lVert\mathbf{k}_3\rVert}}{\lVert\mathbf{k}_2\rVert\, \sin \theta_{23}} & \frac{\frac{\cos \theta_{23}\hspace{.8pt} k_2^2}{\lVert\mathbf{k}_2\rVert}- \frac{k_3^2}{\lVert\mathbf{k}_3\rVert}}{\lVert\mathbf{k}_2\rVert\, \sin \theta_{23}} & \frac{ \frac{\cos \theta_{23}\hspace{.8pt} k_2^3}{\lVert\mathbf{k}_2\rVert}- \frac{k_3^3}{\lVert\mathbf{k}_3\rVert}}{\lVert\mathbf{k}_2\rVert\, \sin \theta_{23}} & 0 & 0 \end{vmatrix}$$
Let us factorize what can be factorized and develop it w.r.t. the third line $$ \begin{split} \Delta_{99} & = \frac{1}{\omega_{\mathbf{k}_3}\,\omega_{\mathbf{k}_1}\,\omega_{\mathbf{k}_2} \lVert\mathbf{k}_1\rVert \sin \theta_{12}\, \lVert\mathbf{k}_2\rVert \sin \theta_{23}}\, \times \Bigg( \\ &\enspace k_3^2\,\begin{vmatrix} - k_1^1 & - k_1^2 & - k_1^3 & - k_1^1 \\ k_2^1 & k_2^2 & k_2^3 & 0 \\ \frac{k_2^1}{\lVert\mathbf{k}_2\rVert} - \frac{\cos \theta_{12}\hspace{.8pt} k_1^1}{\lVert\mathbf{k}_1\rVert} & \frac{k_2^2}{\lVert\mathbf{k}_2\rVert} - \frac{\cos \theta_{12}\hspace{.8pt} k_1^2}{\lVert\mathbf{k}_1\rVert} & \frac{k_2^3}{\lVert\mathbf{k}_2\rVert} - \frac{\cos \theta_{12}\hspace{.8pt} k_1^3}{\lVert\mathbf{k}_1\rVert} & \frac{k_2^1}{\lVert\mathbf{k}_2\rVert} - \frac{ \cos \theta_{12}\hspace{.8pt} k_1^1}{\lVert\mathbf{k}_1\rVert}\\ \frac{\cos \theta_{23}\hspace{.8pt} k_2^1}{\lVert\mathbf{k}_2\rVert}- \frac{k_3^1}{\lVert\mathbf{k}_3\rVert} & \frac{\cos \theta_{23}\hspace{.8pt} k_2^2}{\lVert\mathbf{k}_2\rVert}- \frac{k_3^2}{\lVert\mathbf{k}_3\rVert} & \frac{\cos \theta_{23}\hspace{.8pt} k_2^3}{\lVert\mathbf{k}_2\rVert}- \frac{k_3^3}{\lVert\mathbf{k}_3\rVert} & 0 \end{vmatrix} \\ &\enspace - k_3^1\,\begin{vmatrix} - k_1^1 & - k_1^2 & - k_1^3 & - k_1^2 \\ k_2^1 & k_2^2 & k_2^3 & 0 \\ \frac{k_2^1}{\lVert\mathbf{k}_2\rVert} - \frac{\cos \theta_{12}\hspace{.8pt} k_1^1}{\lVert\mathbf{k}_1\rVert} & \frac{k_2^2}{\lVert\mathbf{k}_2\rVert} - \frac{\cos \theta_{12}\hspace{.8pt} k_1^2}{\lVert\mathbf{k}_1\rVert} & \frac{k_2^3}{\lVert\mathbf{k}_2\rVert} - \frac{\cos \theta_{12}\hspace{.8pt} k_1^3}{\lVert\mathbf{k}_1\rVert} & \frac{k_2^2}{\lVert\mathbf{k}_2\rVert} - \frac{ \cos \theta_{12}\hspace{.8pt} k_1^2}{\lVert\mathbf{k}_1\rVert}\\ \frac{\cos \theta_{23}\hspace{.8pt} k_2^1}{\lVert\mathbf{k}_2\rVert}- \frac{k_3^1}{\lVert\mathbf{k}_3\rVert} & \frac{\cos \theta_{23}\hspace{.8pt} k_2^2}{\lVert\mathbf{k}_2\rVert}- \frac{k_3^2}{\lVert\mathbf{k}_3\rVert} & \frac{\cos \theta_{23}\hspace{.8pt} k_2^3}{\lVert\mathbf{k}_2\rVert}- \frac{k_3^3}{\lVert\mathbf{k}_3\rVert} & 0 \end{vmatrix} \Bigg) \end{split}$$
We use again line operations to cancel what can be cancelled: $$ \begin{split} \Delta_{99} & = \frac{1}{\omega_{\mathbf{k}_3}\,\omega_{\mathbf{k}_1}\,\omega_{\mathbf{k}_2} \lVert\mathbf{k}_1\rVert \sin \theta_{12}\, \lVert\mathbf{k}_2\rVert \sin \theta_{23}}\, \times \\ &\enspace \left( - k_3^2\,\begin{vmatrix} k_1^1 & k_1^2 & k_1^3 & k_1^1 \\ k_2^1 & k_2^2 & k_2^3 & 0 \\ 0 & 0 & 0 & \frac{k_2^1}{\lVert\mathbf{k}_2\rVert} \\ - \frac{k_3^1}{\lVert\mathbf{k}_3\rVert} & - \frac{k_3^2}{\lVert\mathbf{k}_3\rVert} & - \frac{k_3^3}{\lVert\mathbf{k}_3\rVert} & 0 \end{vmatrix} + k_3^1\,\begin{vmatrix} k_1^1 & k_1^2 & k_1^3 & k_1^2 \\ k_2^1 & k_2^2 & k_2^3 & 0 \\ 0 & 0 & 0& \frac{k_2^2}{\lVert\mathbf{k}_2\rVert} \\ - \frac{k_3^1}{\lVert\mathbf{k}_3\rVert} & - \frac{k_3^2}{\lVert\mathbf{k}_3\rVert} & - \frac{k_3^3}{\lVert\mathbf{k}_3\rVert} & 0 \end{vmatrix} \right)\\ & = \frac{1}{\omega_{\mathbf{k}_3}\,\omega_{\mathbf{k}_1}\,\omega_{\mathbf{k}_2} \lVert\mathbf{k}_1\rVert \sin \theta_{12}\, \lVert\mathbf{k}_2\rVert \sin \theta_{23}}\, \left( \frac{- k_3^2\, k_2^1}{\lVert\mathbf{k}_2\rVert \lVert\mathbf{k}_3\rVert} \operatorname{det}(\mathbf{k}_1,\mathbf{k}_2,\mathbf{k}_3) + \frac{ k_3^1\, k_2^2}{\lVert\mathbf{k}_2\rVert \lVert\mathbf{k}_3\rVert} \operatorname{det}(\mathbf{k}_1,\mathbf{k}_2,\mathbf{k}_3) \right) \end{split}$$
I also wonder if one could find a useful interpretation for the factor $\cos \theta_{31}\hspace{.8pt} \frac{k_3^3}{\lVert\mathbf{k}_3\rVert}- \frac{k_1^3}{\lVert\mathbf{k}_1\rVert}$...
2) Inverse: in the plane, if one knows the length of the 3 sides of a triangle than the triangle is known. Setting arbitrarily that one vertex is 0 and one side correspond to the (Ox) axis, one could calculate the sum of the vectors going from (O) to the vertices. We just have to do this on the sphere $S^2$, the role of length being played by the $\theta_{ij}$.
I tried doing this explicitly but it seems really too tedious... Assume $\mathbf{k}_1$ is along the (Oz) axis and $\mathbf{k}_2$ is in the (Ozx) plane. The intersection of the circle with "radius" $\theta_{23}$ around $\frac{\mathbf{k}_2}{\lVert \mathbf{k}_2\rVert}$ with the circle of "radius" $\theta_{31}$ around $\frac{\mathbf{k}_1}{\lVert \mathbf{k}_1\rVert}$ (on the surface of the sphere of radius 1) is given by
\begin{equation} \begin{pmatrix} \cos \theta_{12} & 0 & \sin \theta_{12} \\ 0 & 1 & 0 \\ -\sin \theta_{12} & 0 & \cos \theta_{12} \end{pmatrix} \cdot \begin{pmatrix} r \sin \theta_{23}\, \cos \tau\\ r \sin \theta_{23}\, \sin \tau\\ r \cos \theta_{23} \end{pmatrix} = \begin{pmatrix} r \big(\cos \theta_{12}\, \sin \theta_{23}\, \cos \tau + \sin \theta_{12}\, \cos \theta_{23}\big)\\ r \sin \theta_{23}\, \sin \tau\\ r \big(-\sin \theta_{12}\, \sin \theta_{23}\, \cos \tau + \cos\theta_{12}\, \cos \theta_{23}\big) \end{pmatrix} \overset{!}{=} \begin{pmatrix} r \sin \theta_{31}\, \cos \varphi\\ r \sin \theta_{31}\, \sin \varphi\\ r \cos \theta_{31} \end{pmatrix} \end{equation} I want to find $\varphi$ (azimutal angle as named by wikipedia) as a function of $\big(\theta_{12}, \theta_{23}, \theta_{31}\big)$. Auxiliary variable $\tau \in ]0,\pi[$ determined by the fact the two first component lie in a circle of radius squared $r^2 \sin^2 \theta_{31}$ $ % r\big(\cos \theta_{12}\, \sin \theta_{23}\, \cos \tau + \sin \theta_{12}\, \cos \theta_{23}\big)^2 + \sin^2 \theta_{23}\, \sin^2 \tau \overset{!}{=} \sin^2 \theta_{31} % \cos^2 \theta_{12}\, \sin^2 \theta_{23}\, \cos^2 \tau + 2\, \cos \theta_{12}\, \sin \theta_{23}\, \cos \tau \sin \theta_{12}\, \cos \theta_{23} + \sin^2 \theta_{12}\, \cos^2 \theta_{23} + \sin^2 \theta_{23}\, \sin^2 \tau \overset{!}{=} \sin^2 \theta_{31} $ Carrying on calculation without much care about existence etc. we find $ % \frac{\Delta}{4} &= \cos^2 \theta_{12}\, \sin^2 \theta_{23}\, \sin^2 \theta_{12}\, \cos^2 \theta_{23} - \big(\cos^2 \theta_{12}-1 \big) \sin^2 \theta_{23} \Klammer{\sin^2 \theta_{12}\, \cos^2 \theta_{23} + \sin^2 \theta_{23} - \sin^2 \theta_{31}} \\ % & = \sin^2 \theta_{12}\, \sin^2 \theta_{23} \Klammer{ \cos^2 \theta_{12}\, \cos^2 \theta_{23} - \sin^2 \theta_{12}\, \cos^2 \theta_{23} - \sin^2 \theta_{23} + \sin^2 \theta_{31} } \\ % & = \sin^2 \theta_{12}\, \sin^2 \theta_{23} \Klammer{ \cos \big(2 \theta_{12}\big) \cos^2 \theta_{23} - \sin^2 \theta_{23} + \sin^2 \theta_{31} } $
$$ \cos \tau %& = \frac{-2\, \cos \theta_{12}\, \sin \theta_{23}\, \sin \theta_{12}\, \cos \theta_{23} \pm \sin \theta_{12}\, \sin \theta_{23} \sqrt{ \cos \big(2 \theta_{12}\big) \cos^2 \theta_{23} - \sin^2 \theta_{23} + \sin^2 \theta_{31} }}{2 \, \sin^2 \theta_{12} \, \sin^2 \theta_{23}} = \frac{-2\, \cos \theta_{12}\, \cos \theta_{23} \pm \sqrt{ \cos \big(2 \theta_{12}\big) \cos^2 \theta_{23} - \sin^2 \theta_{23} + \sin^2 \theta_{31} }}{2 \, \sin \theta_{12} \, \sin \theta_{23}} $$ ...
Assuming that we have obtained our three vertices $\frac{\tilde{\mathbf{k}}_i}{\lVert\tilde{\mathbf{k}}_i\rVert}$, we have to multiply them by their norm (given by the $\omega_{\mathbf{k}}$) and determine the rotation that brings $\tilde{\mathbf{k}}_1 + \tilde{\mathbf{k}}_2 + \tilde{\mathbf{k}}_3$ to the sum which was denoted $\mathbf{k}$...
Ok, I stop here, it becomes nonsense