Circles $\omega_1$, $\omega_2$, and $\omega_3$ are centered at $M$, $N$, and $O$, respectively.  The points of tangency between $\omega_2$ and $\omega_3$, $\omega_3$ and $\omega_1$, and $\omega_1$ and $\omega_2$ are tangent at $A$, $B$, and $C$, respectively.  Line $MO$ intersects $\omega_3$ and $\omega_1$ again at $P$ and $Q$ respectively, and line $AP$ intersects $\omega_2$ again at $R$.  Given that $ABC$ is an equilateral triangle of side length $1$, compute the area of $PQR$.
1. **Identify the Geometry and Given Information:**
   - Circles $\omega_1$, $\omega_2$, and $\omega_3$ are centered at $M$, $N$, and $O$ respectively.
   - Points of tangency between $\omega_2$ and $\omega_3$, $\omega_3$ and $\omega_1$, and $\omega_1$ and $\omega_2$ are $A$, $B$, and $C$ respectively.
   - Line $MO$ intersects $\omega_3$ and $\omega_1$ again at $P$ and $Q$ respectively.
   - Line $AP$ intersects $\omega_2$ again at $R$.
   - $\triangle ABC$ is an equilateral triangle with side length $1$.

2. **Determine the Geometry of $\triangle ONM$:**
   - Since $\triangle ABC$ is equilateral and the points of tangency form an equilateral triangle, the centers $M$, $N$, and $O$ must also form an equilateral triangle.
   - The side length of $\triangle ONM$ is twice the radius of the circles, which is $2$.

3. **Analyze $\triangle BPA$:**
   - $\angle BPA = \frac{1}{2} \angle BOA = 30^\circ$ because $\angle BOA = 60^\circ$ (since $\triangle ONM$ is equilateral).
   - $\triangle BPA$ is a 30-60-90 triangle.
   - In a 30-60-90 triangle, the sides are in the ratio $1 : \sqrt{3} : 2$.
   - Given $BP = 1$, we have $AP = \sqrt{3}$.

4. **Calculate the Area of $\triangle APB$:**
   - The area of $\triangle APB$ is given by:
     \[
     [\triangle APB] = \frac{1}{2} \times AP \times BP \times \sin(30^\circ) = \frac{1}{2} \times \sqrt{3} \times 1 \times \frac{1}{2} = \frac{\sqrt{3}}{4}
     \]

5. **Determine the Midpoints and Area of $\triangle PQR$:**
   - $A$ is the midpoint of $PR$, and $B$ is the midpoint of $PQ$.
   - Since $A$ and $B$ are midpoints, $PR = 2 \times AP = 2\sqrt{3}$ and $PQ = 2 \times BP = 2$.
   - The area of $\triangle PQR$ can be calculated using the fact that it is four times the area of $\triangle APB$:
     \[
     [\triangle PQR] = 4 \times [\triangle APB] = 4 \times \frac{\sqrt{3}}{4} = \sqrt{3}
     \]

The final answer is $\boxed{\sqrt{3}}$