Let $\ell$ be a line in the plane and let $90^\circ<\theta<180^\circ$. Consider any distinct points $P,Q,R$ that satisfy the following: 

(i) $P$ lies on $\ell$ and $PQ$ is perpendicular to $\ell$

(ii) $R$ lies on the same side of $\ell$ as $Q$, and $R$ doesn’t lie on $\ell$

(iii) for any points $A,B$ on $\ell$, if $\angle ARB=\theta$ then $\angle AQB \geq \theta$. 

Find the minimum value of $\angle PQR$.
1. **Understanding the problem and setting up the scenario:**
   - Let $\ell$ be a line in the plane.
   - Let $P$ be a point on $\ell$.
   - Let $Q$ be a point such that $PQ$ is perpendicular to $\ell$.
   - Let $R$ be a point on the same side of $\ell$ as $Q$ and not on $\ell$.
   - For any points $A, B$ on $\ell$, if $\angle ARB = \theta$, then $\angle AQB \geq \theta$.

2. **Claim:**
   - We claim that the minimum value of $\angle PQR$ is $270^\circ - \theta$.

3. **Proof by contradiction:**
   - Suppose $\angle PQR < 270^\circ - \theta$.
   - Consider a point $X$ on $\ell$ such that $X$ is on the opposite side of $PQ$ as $R$, $\triangle PXR$ does not contain $Q$, and $\angle PXR < 180^\circ - \theta$.
   - This is possible by making line $RX$ arbitrarily close to $Q$.
   - Let $\angle PXR = 180^\circ - \theta - 2\alpha$.
   - Consider the point $A$ such that $\angle PAR = 180^\circ - \theta - \alpha$ and $B$ on the other side such that $\angle RBA = \alpha$.
   - The key point is that $XR$ is tangent to $(ARB)$, and $\angle ARB = \theta$.
   - By construction, $\triangle XRB$ does not contain $Q$.
   - Thus, $Q$ lies outside $(ARB)$ on the same side of $AB$ as $R$, which means $\angle ARB > \angle AQB$.
   - This violates the condition that $\angle AQB \geq \theta$.

4. **Verification of the claim:**
   - Now take $R$ such that $\angle PQR = 270^\circ - \theta$.
   - Suppose $A, B$ exist such that $\angle AQB < \theta$.
   - Let $RQ$ meet $\ell$ at $Y$.
   - Then, without loss of generality, if $A$ is on the opposite side of $PQ$ than $R$, then $A$ must lie on the opposite side of $Y$ than $P$ (else $B$ cannot exist).
   - Thus, triangle $ARB$ must contain $Q$, which means $\angle AQB > \angle ARB$, a contradiction.

Therefore, the minimum value of $\angle PQR$ is $270^\circ - \theta$.

The final answer is $\boxed{270^\circ - \theta}$