Consider the unit vector of
$$ \check{n} = \begin{bmatrix} \sin \phi \sin\gamma \\ \cos \phi\\ \sin \phi \cos\gamma \end{bmatrix} \, . \tag{1} $$
Fig.1 - Image courtesy of dynref.engr.illinois.edu/rvs.html
From here the angular velocity of a unit vector should be:
$$ \vec{\omega} = \check{n} \times \frac{d\check{n}}{dt} \, , \tag{2} $$
and consequently
$$ \Rightarrow \vec{\omega} = \begin{bmatrix} \sin \phi \sin\gamma \\ \cos \phi\\ \sin \phi \cos\gamma \end{bmatrix} \times \begin{bmatrix} \dot{\phi} \cos \phi \sin \gamma + \dot{\gamma} \sin \phi \cos \gamma \\ -\dot{\phi} \sin \phi\\ \dot{\phi} \cos \phi \cos \gamma - \dot{\gamma} \sin \phi \sin\gamma \end{bmatrix} \, . \tag{3} $$
The same page has stated that the relative angular velocity is addable
$$\vec{\omega}_{13} = \vec{\omega}_{12} + \vec{\omega}_{23} \, . \tag{4}$$
Therefore, I should be able to write
$$ \vec{\omega} = \begin{bmatrix} \dot{\phi} \cos \gamma \\ 0\\ -\dot{\phi} \sin \gamma \end{bmatrix} + \begin{bmatrix} 0 \\ \dot{\gamma} \\ 0 \end{bmatrix} = \begin{bmatrix} \dot{\phi} \cos \gamma \\ \dot{\gamma} \\ -\dot{\phi} \sin \gamma \end{bmatrix} \, . \tag{5} $$
However, I don't know how to get from Eq.3 to Eq.5 and I have the suspicion that there is something wrong. I would appreciate if you could help me know if 3 and 5 are identical and if yes how to get to 5 from 3. If not which one is correct then and where is my mistake?
P.S.1. The best simplification I can get from Eq.3 is
$$ \begin{bmatrix} \dot{\phi} \cos \gamma - \dot{\gamma} \sin \phi \cos \phi \sin \gamma \\ \dot{\gamma} \sin^2 \phi\\ -\dot{\phi} \sin \gamma - \dot{\gamma} \sin \phi \cos \phi \cos \gamma \end{bmatrix} \, , \tag{6} $$
which most certainly doesn't seem identical to Eq.5!