Given a circle of radius $r$ and a tangent line $\ell$ to the circle through a given point $P$ on the circle.  From a variable point $R$ on the circle, a perpendicular $RQ$ is drawn to $\ell$ with $Q$ on $\ell$.  Determine the maximum of the area of triangle $PQR$.
1. **Setup and Notation:**
   - Let \( w = C(O, r) \) be a circle with center \( O \) and radius \( r \).
   - Let \( \ell \) be a tangent line to the circle at point \( P \).
   - Let \( R \) be a variable point on the circle, and \( Q \) be the foot of the perpendicular from \( R \) to \( \ell \).

2. **Expression for the Area of \(\triangle PQR\):**
   - Denote \( m(\widehat{POR}) = 2\phi \), where \( 0 \leq \phi \leq \frac{\pi}{2} \).
   - The length \( PR = 2r \sin \phi \) (since \( PR \) is the chord subtending angle \( 2\phi \) at the center).
   - The length \( PQ = 2r \sin \phi \cos \phi \) (since \( PQ \) is the projection of \( PR \) on the tangent line).
   - The area of \(\triangle PQR\) is given by:
     \[
     [PQR] = \frac{1}{2} \cdot PR \cdot PQ \cdot \sin \phi = \frac{1}{2} \cdot 2r \sin \phi \cdot 2r \sin \phi \cos \phi \cdot \sin \phi = 2r^2 \sin^3 \phi \cos \phi
     \]

3. **Maximizing the Area:**
   - We need to maximize \( 2r^2 \sin^3 \phi \cos \phi \).
   - Let \( f(\phi) = \sin^3 \phi \cos \phi \).
   - To find the maximum, we take the derivative and set it to zero:
     \[
     f'(\phi) = 3 \sin^2 \phi \cos^2 \phi - \sin^4 \phi = \sin^2 \phi \cos \phi (3 \cos \phi - \sin \phi)
     \]
     \[
     f'(\phi) = 0 \implies \sin^2 \phi \cos \phi (3 \cos \phi - \sin \phi) = 0
     \]
     - This gives us critical points at \( \phi = 0 \), \( \phi = \frac{\pi}{2} \), and \( 3 \cos \phi = \sin \phi \).
     - Solving \( 3 \cos \phi = \sin \phi \):
       \[
       \tan \phi = 3 \implies \phi = \tan^{-1}(3)
       \]
     - However, we need to check within the interval \( [0, \frac{\pi}{2}] \).

4. **Verification of Maximum:**
   - We also need to check the boundary values:
     \[
     f(0) = 0, \quad f\left(\frac{\pi}{2}\right) = 0
     \]
   - The maximum value occurs at \( \phi = \frac{\pi}{6} \):
     \[
     f\left(\frac{\pi}{6}\right) = \sin^3\left(\frac{\pi}{6}\right) \cos\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^3 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{16}
     \]
   - Therefore, the maximum area is:
     \[
     [PQR] = 2r^2 \cdot \frac{\sqrt{3}}{16} = \frac{r^2 \sqrt{3}}{8}
     \]

The final answer is \(\boxed{\frac{r^2 \sqrt{3}}{8}}\).