If we express the change in anglular velocity as $\Delta\vec\omega$ in local coordinates, with for example the angles $\phi$, $\theta$ and $\psi$ rotations about the $Z$, $X$ and $Z$ respectively then the answer is
$$\Delta\vec{\omega}=\dot{\phi}\hat{k}+\mathrm{{Rot}(\hat{k},\phi})\left(\dot{\theta}\hat{i}+\mathrm{{Rot}(\hat{i},\theta})\left(\dot{\psi}\hat{k}\right)\right)$$
Where Rot(axis,angle) is a 3x3 rotation matrix. We apply the angular rotation components in sequence on the local coordinate axes this way. Also $\hat{i}=(1,0,0)$, $\hat{j}=(0,1,0)$ and $\hat{k}=(0,0,1)$.
Line of node must be the common normal between the two $z$-axes. We typically denote that as the $x$-axis (see Denavit-Hartenberg notation).
This is a problem I've struggled with before. The issue is that most textbooks are not clear enough in their definitions of the fixed and rotating reference frames. I've tried to be as comprehensive as possible in my answer, I hope it helps!
To describe the motion of a rigid body through space we must make use of two separate reference frames:
Firstly, a fixed (i.e. inertial) reference frame with a set of orthonormal basis vectors $\mathbf{e}_i$. These vectors are constant in time to all inertial observers. I will call the inertial reference frame $K$, and this set of vectors $S$.
Secondly, the body frame - a non-inertial reference frame which moves with the rigid body. A second set of orthonormal basis vectors $\mathbf{e}'_j$ move with this frame. I will call this frame of reference $K'$, and this set of vectors $S'$.
The diagram below illustrates this picture (the weird blob is the rigid body!).
Any vector in space can be written as a linear combination of either set of basis vectors: we can write
$$
\mathbf{A} = A_i\mathbf{e}_i = A'_j\mathbf{e}'_j.
$$
In particular, if the vector $\mathbf{A}$ is fixed to the rigid body, then the components $A_i$ are functions of time, but, since the projection of $\mathbf{A}$ onto each of the $\mathbf{e}'_j$ is constant, the $A'_j$ are constant. This holds in both $K$ and $K'$.
Importantly, an observer from either reference frame can use either set of basis vectors. The difference is in how they view the behaviour of each set of vectors: from the perspective of $K$, the basis vectors $\mathbf{e}_i$ are fixed, but the basis vectors $\mathbf{e}'_j$ are functions of time; however, to an observer in $K'$, i.e. from the perspective of the rigid body, the $\mathbf{e}'_j$ are fixed, and it is instead the vectors $\mathbf{e}_i$ which are time dependent.
Thus, when observed from $K'$,
$$
\left(\frac{\text{d}}{\text{d}t}\right)_{K'}\mathbf{e}'_i = \mathbf{0}
$$
so the rate of change of a vector $\mathbf{A} = A'_i\mathbf{e}'_i$ appears to be
$$
\left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K'} = \dot{A'}_i\mathbf{e}'_i.
$$
In $S$ howver, the $\mathbf{e}'_i$ basis vectors, are functions of time. I show below that, as observed from $K$,
$$
\left(\frac{\text{d}}{\text{d}t}\right)_{K}\mathbf{e}'_i = \boldsymbol{\omega}\times\mathbf{e}'_i\tag{1}
$$
for some time-dependent vector $\boldsymbol{\omega}$, so
\begin{align}
\left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K} &= \dot{A'}_i\mathbf{e}'_i +A'_i\dot{\mathbf{e}}'_i\\
& = \dot{A'}_i\mathbf{e}'_i + A'_i \boldsymbol{\omega}\times\mathbf{e}'_i\\
& = \dot{A'}_i\mathbf{e}'_i+ \boldsymbol{\omega}\times(A'_i \mathbf{e}'_i)\\
& = \dot{A'}_i\mathbf{e}'_i +\boldsymbol{\omega}\times\mathbf{A},
\end{align}
and thus
$$
\left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K'} = \left(\frac{\text{d}\mathbf{A}}{\text{d}t}\right)_{K} - \boldsymbol{\omega}\times\mathbf{A}. \tag{2}
$$
This is the crux of the problem: to an observer in the moving frame, every vector appears to have an additional term in its time derivative, when in fact it is the frame itself that is changing.
With this, Euler's equations follow quite naturally. Let $\mathbf{T}$ be the torque on the body, and let $\underline{\mathbf{I}}\boldsymbol{\omega}$ be the angular momentum of the body, where $\underline{\mathbf{I}}$ is the moment of inertia tensor. The key point is that as measured in $K'$, the moment of inertia tensor is constant. (This is not true in $K$, as the distribution of the mass of the body with respect to the $\mathbf{e}_i$ basis changes with time.) Using equation (2), this means that
$$
\left(\frac{\text{d}}{\text{d}t}\right)_{K'}\underline{\mathbf{I}}\boldsymbol{\omega} = \underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K'}\boldsymbol{\omega} = \underline{\mathbf{I}}\left[\left(\frac{\text{d}}{\text{d}t}\right)_{K}\boldsymbol{\omega}-\underbrace{\boldsymbol{\omega}\times\boldsymbol{\omega}}_{\mathbf{0}}\right] = \underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K}\boldsymbol{\omega}.\tag{$\star$}
$$
Finally, the rate of change of angular momentum in the body is equal to the applied torque:
$$
\mathbf{T}= \left(\frac{\text{d}}{\text{d}t}\right)_{K}\underline{\mathbf{I}}\boldsymbol{\omega} = \left(\frac{\text{d}}{\text{d}t}\right)_{K'}\underline{\mathbf{I}}\boldsymbol{\omega}+\boldsymbol{\omega}\times(\underline{\mathbf{I}}\boldsymbol{\omega}) =\underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K}\boldsymbol{\omega} +\boldsymbol{\omega}\times(\underline{\mathbf{I}}\boldsymbol{\omega}).
$$
This is Euler's equation! It holds from the perspective of the inertial frame, but is derived using the basis vectors of the non-inertial frame.
All that remains is to prove equation (1). To do so, it is best to forget about the body frame $K'$, and to think of the $\mathbf{e}'_i$ as simply a set of vectors which move with respect to the $\mathbf{e}_i$, as seen from $K$. The moving basis vectors $\mathbf{e}'_j(t)$ are related to the fixed basis vectors $\mathbf{e}_i$ by a time dependent rigid rotation, which can be described by a matrix $\underline{\mathbf{R}}(t)$. Using Einstein notation, we have
$$
\mathbf{e}'_i(t) = R_{ij}(t) \mathbf{e}_j, \qquad\text{and}\qquad \mathbf{e}_i = R_{ij}^T(t) \mathbf{e}'_j(t). \tag{1}
$$
Since $\underline{\mathbf{R}}$ is a rotation matrix, it is orthogonal, so for all times $t$ it satisfies
$$
\underline{\mathbf{R}}(t)\underline{\mathbf{R}}^T(t) = \underline{\mathbf{I}},
$$
where $\underline{\mathbf{I}}$ is the identity matrix. Differentiating this with respect to time we find
$$
\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t)+\underline{\mathbf{R}}(t)\underline{\dot{\mathbf{R}}}^T(t)= \underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t) + \left(\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t)\right)^T = \underline{\mathbf{0}},
$$
so the matrix $\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t)$ must be antisymmetric. This implies that there exists some vector $\boldsymbol{\omega} = (\omega_1, \omega_2, \omega_3)$ such that
$$
\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t) = [\boldsymbol{\omega}\times]
$$
where the matrix
$$
[\boldsymbol{\omega}\times] = \begin{pmatrix}0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 &0 \end{pmatrix}
$$
acts on vectors like a cross product as $[\boldsymbol{\omega}\times]\mathbf{A} = \boldsymbol{\omega}\times\mathbf{A}$ (see here if you've not seen this before). For our purposes, it will be most useful to know the components of this matrix in the moving basis $\mathbf{e}'_j$. Using $\boldsymbol{\omega} = \omega'_j\mathbf{e}'_j$, we have
$$
\left(\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t)\right)'_{ij} = \mathbf{e}'^T_i\left(\underline{\dot{\mathbf{R}}}(t)\underline{\mathbf{R}}^T(t)\right)\mathbf{e}'_j = \epsilon_{ijk}\omega'_k.
$$
With this, we can calculate the time derivatives of the primed basis vectors:
\begin{align}
\dot{\mathbf{e}}'_i(t) &= \dot{R}_{ij}(t) \mathbf{e}_j \\
&= \dot{R}_{ij} R^T_{jk}\mathbf{e}'_k\\
&= \epsilon_{ikm}\omega'_m \mathbf{e}'_k\\
&= \epsilon_{ikm}(\boldsymbol{\omega}\cdot\mathbf{e}'_m)\mathbf{e}'_k\\
&= \frac{1}{2} \epsilon_{ikm}(\boldsymbol{\omega}\cdot\mathbf{e}'_m)\mathbf{e}'_k - \frac{1}{2}\epsilon_{ikm}(\boldsymbol{\omega}\cdot\mathbf{e}'_k)\mathbf{e}'_m\\
&=\frac{1}{2}\epsilon_{ikm}\boldsymbol{\omega}\times(\mathbf{e}'_k\times\mathbf{e}'_m)\\
&=\frac{1}{2}\epsilon_{ikm}\boldsymbol{\omega}\times(\epsilon_{kml}\mathbf{e}'_l)\\
&=\delta_il\boldsymbol{\omega}\times\mathbf{e}'_l\\
&=\boldsymbol{\omega}\times\mathbf{e}'_i(t).\tag{2}
\end{align}
Thus, as seen from the inertial frame, there exists an instantaneous angular velocity vector $\boldsymbol{\omega}$ around which the primed basis vectors $\mathbf{e}'_i(t)$ rotate.
Finally, here is an alternative derivation of Euler's equation that I am particularly fond of. We proceed by directly by differentiating the moment of inertia tensor. To do this, we align the $\mathbf{e}'_i$ with the principal axes of the body. We denote the principal moment of the body along the $\mathbf{e}'_i$ axis as $I_i$. Since the body is rigid, the pricipal moments are constants, and so we can explicitly write the moment of inertial tensor as
$$
\underline{\mathbf{I}} = \sum_i I_i \mathbf{e}'_i \otimes \mathbf{e}'_i
$$
(summation convention not used here).
The time derivative of $\underline{\mathbf{I}}$ (in $K$) is then
$$
\left(\frac{\text{d}}{\text{d}t}\right)_{K} \underline{\mathbf{I}} = \sum_i I_i\left[\dot{\mathbf{e}}'_i\otimes\mathbf{e}'_i+\mathbf{e}'_i\otimes\dot{\mathbf{e}}'_i\right] = \sum_i I_i\left[(\boldsymbol{\omega}\times\mathbf{e}'_i)\otimes\mathbf{e}'_i+\mathbf{e}'_i\otimes(\boldsymbol{\omega}\times\mathbf{e}'_i)\right].
$$
Now, let $\mathbf{b}$ and $\mathbf{c}$ be arbitrary vectors, and $\underline{\mathbf{A}}$ be a linear operator. Using the definition of the tensor product, we have
$$
\left[(\underline{\mathbf{A}}\mathbf{b})\otimes\mathbf{c}\right]_{ij}=\left(\underline{\mathbf{A}}\mathbf{b}\right)_i c_j = A_{ik}b_k c_j = \left(\underline{\mathbf{A}}\right)_{ik}\left(\mathbf{b}\otimes\mathbf{c}\right)_{kj} = \left[\underline{\mathbf{A}}(\mathbf{b}\otimes\mathbf{c})\right]_{ij}
$$
and
$$
\left[\mathbf{b}\otimes(\underline{\mathbf{A}}\mathbf{c})\right]_{ij}=b_i\left(\underline{\mathbf{A}}\mathbf{c}\right)_j = b_i A_{jk} c_k = \left(\mathbf{b}\otimes\mathbf{c}\right)_{ik}\left(\underline{\mathbf{A}}^T\right)_{kj} = \left[(\mathbf{b}\otimes\mathbf{c})\underline{\mathbf{A}}^T\right]_{ij}.
$$
Therefore,
\begin{align}
\left(\frac{\text{d}}{\text{d}t}\right)_{K} \underline{\mathbf{I}}&= \sum_i I_i\left[(\boldsymbol{\omega}\times\mathbf{e}'_i)\otimes\mathbf{e}'_i+\mathbf{e}'_i\otimes(\boldsymbol{\omega}\times\mathbf{e}'_i)\right] \\
&= \sum_i I_i\left[[\boldsymbol{\omega}\times](\mathbf{e}'_i\otimes\mathbf{e}'_i)+(\mathbf{e}'_i\otimes\mathbf{e}'_i)[\boldsymbol{\omega}\times]^T\right]\\
&=[\boldsymbol{\omega}\times]\underline{\mathbf{I}}-\underline{\mathbf{I}}[\boldsymbol{\omega}\times]
\end{align}
where $[\boldsymbol{\omega}\times]$ is the operator defined in the previous section, and in the last line I have used that this operator is anti-symmetric.
Thus, the time derivative of the angular momentum is
\begin{align}
\mathbf{T} &= \left(\frac{\text{d}}{\text{d}t}\right)_{K} (\underline{\mathbf{I}}\boldsymbol{\omega}) \\
&= \underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \boldsymbol{\omega}+\boldsymbol{\omega}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \underline{\mathbf{I}}\\
&=\underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \boldsymbol{\omega} + [\boldsymbol{\omega}\times]\underline{\mathbf{I}}\boldsymbol{\omega}-\underline{\mathbf{I}}\underbrace{[\boldsymbol{\omega}\times]\boldsymbol{\omega}}_{\boldsymbol{\omega}\times\boldsymbol{\omega}=\mathbf{0}}\\
&=\underline{\mathbf{I}}\left(\frac{\text{d}}{\text{d}t}\right)_{K} \boldsymbol{\omega} + \boldsymbol{\omega}\times(\underline{\mathbf{I}}\boldsymbol{\omega})
\end{align}
as before.
Best Answer
Any motion of a rigid object can be decomposed into translation (straight-line motion) of the center of mass, and rotation around the center of mass. In a movie scene, an asteroid would probably be both rotating and translating with respect to the camera, because it makes the motion look more complicated that way. But if you imagine a different camera moving along with the asteroid, it would see the motion as only rotation.
Now, any rotation can always be expressed as a rotation around a fixed axis with a fixed speed. Even if it appears that the asteroid is rotating around several different axes simultaneously, there really is just one axis, according to Euler's theorem. In movie scenes, they tend to orient that axis at an odd angle to the camera, again to make it look more complicated, but given enough information about the characteristics of the motion, you can always find the one axis of rotation and identify the speed of the rotation around it.