Let $\omega$ be the incircle of a fixed equilateral triangle $ABC$. The line $\ell$ is tangent to $\omega$ and intersects the interior of segments $BC$ and $CA$ at points $P$ and $Q$ respectively. Point $R$ is chosen such that $PR = PA$ and $QR = QB$. We need to find all possible locations of point $R$ for all choices of the line $\ell$.

To solve this question, consider the following steps:

1. Basic Setup and Geometry:
The incircle $\omega$ of triangle $ABC$ is tangent to sides $BC$, $CA$, and $AB$. Since $ABC$ is equilateral, the incircle is symmetric with respect to the perpendicular bisectors of the sides.

2. Properties of Points on Elliptical Paths:
Since $\ell$ is tangent to $\omega$, the distances $PA$ and $PB$ are equal (due to tangency). Similarly, $QA = QB$. Thus, $R$ satisfies the conditions $PR = PA$ and $QR = QB$. Point $R$ must lie on the locus where these equalities can hold.

3. Symmetry and Locus Characterization:
According to the given equalities, $R$ can be considered as being equidistant from points $A$ and $B$. To maintain these equal distances as $\ell$ varies, $R$ must lie on a line that preserves these symmetries.

4. Identifying the Locus:
It can be observed that the conditions $PR = PA$ and $QR = QB$ are satisfied if and only if $R$ lies on a line equidistant from $A$ and $B$. This geometric locus is the perpendicular bisector of segment $BC$.

Thus, the locus of all possible points $R$, making $PR = PA$ and $QR = QB$ true for all choices of line $\ell$, is given by the perpendicular bisector of segment $BC$.

Therefore, the final answer is:
$$\boxed{\text{The perpendicular bisector of segment } BC}.$$