Let $A$ and $X(t)$ be the two matrices
$$ A = \begin{bmatrix}
0 & -\omega_3 & \omega_2 \\
\omega_3 & 0 &-\omega_1 \\
-\omega_2 & \omega_1 & 0 \end{bmatrix}
\quad\text{ and }\quad
X(t) = \begin{bmatrix} x_1 (t) \\ x_2 (t) \\ x_3 (t) \end{bmatrix}$$
and $\omega = \sqrt{\omega_1^2 + \omega_2^2 + \omega_3^2} \ge 0$. When $A$ is constant,
the ODE for $X(t)$
$$\frac{d}{dt}X(t) = A X(t)$$
has solution of the form
$$X(t) = e^{tA} X(0)\quad\text{ where }\quad
e^{tA} = \sum_{k=0}^\infty \frac{t^k}{k!} A^k\tag{*1}
$$
When $A$ is a skew symmetric matrix, it satisfy an interesting identity:
$$A^3 = -\omega^2 A$$
A consequence of this is for any even function $f(z)$ whose power series expansion converges rapidly enough, you will find
$$f(A)A = f(i\omega)A\quad\text{ and }\quad f(A)A^2 = f(i\omega)A^2$$
Apply this to $e^{tA}$, one can convert the series into a quadratic polynomial in $A$!
$$\begin{align}
e^{tA}
&= I_n + \left(\sum_{k=0}^\infty\frac{t^{2k+1}}{(2k+1)!}(A^2)^k\right)A
+ \left(\sum_{k=0}^\infty\frac{t^{2k+2}}{(2k+2)!}(A^2)^k\right)A^2\\
&= I_n + \left(\sum_{k=0}^\infty\frac{t^{2k+1}}{(2k+1)!}(i\omega)^{2k}\right)A
+ \left(\sum_{k=0}^\infty\frac{t^{2k+2}}{(2k+2)!}(i\omega)^{2k}\right)A^2\\
&= I_n + \frac{\sin(t\omega)}{\omega} A + \frac{1-\cos(t\omega)}{\omega^2} A^2
\end{align}\tag{*2}
$$
Please note that during practical calculation, there is no need to manipulate power series explicitly like this. Instead, one can formally manipulate $(*2)$ as follows
$$
e^{tA}
= I_n + \frac{\sinh(tA)}{A} A + \frac{\cosh(tA) - I_n}{A^2} A^2
= I_n + \frac{\sin(t\omega)}{\omega} A + \frac{1-\cos(t\omega)}{\omega^2} A^2
$$
Please keep in mind in above manipulation, expression like $\displaystyle\;\frac{\sinh(tA)}{A}\;$ doesn't
mean compute $\sinh(tA)$ by a power series expansion and then divide it by $A$. Instead,
it means evaluate the power series associated with $\displaystyle\;\frac{\sin(tz)}{z}$ "at" $z = A$.
Finally, let us rephrase the solution $(*1)$ in terms of vectors. Let
$\hat{e}_i, i = 1,2,3$ be the canonical basis of $\mathbb{R}^3$ as a vector space.
Let $\vec{\omega}$ and $\vec{X}(t)$ be the
two vectors
$$
\vec{\omega} = \omega_1 \hat{e}_1 + \omega_2 \hat{e}_2 + \omega_3 \hat{e}_3
\quad\text{ and }\quad
\vec{X}(t) = x_1(t) \hat{e}_1 + x_2(t) \hat{e}_2 + x_3(t) \hat{e}_3
$$
Let $\hat{\omega}$ be the unit vector $\displaystyle\;\frac{\vec{\omega}}{\omega}\;$.
As mentioned in the question, application of $A$ to the column vector $X(t)$ is equivalent to a cross product between $\vec{\omega}$ and $\vec{X}(t)$.
$$A X(t)\quad\leftrightarrow\quad \vec{\omega} \times \vec{X}(t)$$
If we substitute $(*2)$ into $(*1)$ and employ this type of correspondence between matrices and vectors. We will obtain
$$
\vec{X}(t)
= \vec{X}(0) + \sin(t\omega)\big( \hat{\omega} \times \vec{X}(0) \big) +
(1 - \cos(t\omega)) \big( \hat{\omega} \times ( \hat{\omega} \times \vec{X}(0)) \big)
$$
If we split $\vec{X}(0)$ into two orthogonal vectors, one parallel and another perpendicular to $\hat{\omega}$:
$$\vec{X}(0) = \vec{X}_{\parallel}(0) + \vec{X}_{\perp}(0)
\quad\text{ where }\quad
\begin{cases}
\vec{X}_{\parallel}(0) &= ( \vec{X}_(0) \cdot \hat{\omega} ) \hat{\omega}\\
\vec{X}_{\perp}(0) &= -\hat{\omega} \times ( \hat{\omega} \times \vec{X}(0) )
\end{cases}
$$
We can re-express $\vec{X}(t)$ in a much more informative form:
$$\vec{X}(t) = \underbrace{\vec{X}_{\parallel}(0)}_{\vec{X}_{\parallel}(t)} +
\underbrace{\cos(t\omega)\vec{X}_{\perp}(0) + \sin(t\omega)\big(\hat{\omega} \times \vec{X}_{\perp}(0) \big)}_{\vec{X}_{\perp}(t)}$$
As one can see, the parallel part of $\vec{X}(t)$ remains a constant
while the perpendicular part of $\vec{X}(t)$ rotate around the axis in
direction of $\hat{\omega}$ in circle.