Let's first consider the case when the radii of $\omega_2$ and $\omega_3$ are different (without loss of generality assume that the radius of $\omega_3$ is greater than the radius of $\omega_2$). Consider three homotheties: $l_1$ centered at $Q$ and mapping $\omega_3$ to $\omega_1$; $l_2$ centered $P$ and mapping $\omega_1$ to $\omega_2$; and $l_3 = l_2 \circ l_1$ mapping $\omega_3$ to $\omega_2$. Since $l_1$ maps $A$ to $C$, $l_2$ maps $C$ to $B$, then $l_3$ maps $A$ to $B$, so its center (denoted by $I$) lies on the line $AB$.
Denote $\angle ACB = \angle PCQ = \alpha$, this angle is fixed (by a fixed value we denote any value which doesn't depend on the position of $C$). Then the angle $\angle AOB = 2\alpha$ is also fixed, which means that all triangles $AOB$ are similar to each other and the ratio $\frac{OA}{AB} = \frac{1}{2\sin\alpha}$ is fixed.
Let $\frac{IA}{IB} = t > 1$ be the homothety coefficient of $l_3$, this number is fixed. Then the ratio
$$\frac{AB}{IA} = \frac{IB-IA}{IA} = \frac{1}{t} - 1 \text{ is also fixed.}$$
In the triangle $IAO$ we can find the ratio
$$\frac{OA}{IA} = \frac{OA}{AB} \cdot \frac{AB}{IA} = \frac{1}{2\sin\alpha} \cdot \left(\frac{1}{t} - 1\right)$$
and the angle
$$\angle OAI = 180^{\circ} - \angle OAB = 90^{\circ} + \alpha.$$
Hence all triangles $IAO$ are similar to each other, in particular, the angle $AIO$ and the ratio $\frac{IO}{IA}$ are fixed. Therefore for any position of the point $C$ on the circle $\omega_1$, the spiral similarity $\ell$ with center $I$, angle $AIO$ and coefficient $\frac{IO}{IA}$ maps $A$ to $O$. Since the point $A$ varies over all positions on $\omega_3$, the locus of $O$ is the circle $\omega = \ell(\omega_3)$.
Note that the homothety $\ell_3$ and the spiral similarity $\ell$ have a common center $I$. Therefore $\ell(B) = \ell \circ \ell_3(A) = \ell_3 \circ \ell(A) = \ell_3(O_3) = O_2$. Hence $\ell$ maps the triangle $AOB$ to the triangle $O_3O'O_2$ and
$$\angle O_3O'O_2 = \angle AOB = 2\angle ACB = \angle QO_1P = \angle O_3O_1O_2,$$
from which we conclude that $O'$ lies on the circumcircle of the triangle $O_1O_2O_3$.

Let now the radii of the circles $\omega_2$ and $\omega_3$ be equal. In this case $\ell_3$ is a translation by the vector $\overrightarrow{O_3O_2}$, the point $I$ is not defined and the quadrilateral $ABO_2O_3$ is a parallelogram. Let $D$ be a such point that triangles $ABD$ and $O_3O_2O_1$ are equal and have the same orientation. The quadrilateral $O_3ADO_1$ is a parallelogram, while the triangles $QAO_3$ and $QCO_1$ are similar. Hence the segments $DO_1$ and $O_1C$ are parallel and
$$DO_1 + O_1C = O_3A \left(1 + \frac{QO_1}{QO_3}\right) = O_3Q \left(1 + \frac{QO_1}{QO_3}\right) = O_1O_3 = DA = DB.$$