1. **State the problem:** We have a circle with center $O$ and points $P$, $Q$, and $R$ on the circumference. The arc $PQR$ has length 14 and radius $r$. The angle subtended by arc $PQ$ at the center $O$ is $\frac{2\pi}{3}$ radians. The angle $\angle POR$ is a right angle, i.e., $\frac{\pi}{2}$ radians. We need to find the length of arc $QR$. 2. **Recall the formula for arc length:** $$\text{Arc length} = r \times \text{central angle in radians}$$ 3. **Find the central angle for arc $PQR$:** Since $PQR$ is an arc from $P$ to $R$ passing through $Q$, the central angle $\angle POR$ corresponds to the arc $PQR$. Given $\angle POR = \frac{\pi}{2}$ radians. 4. **Use the arc length formula for arc $PQR$:** $$14 = r \times \frac{\pi}{2}$$ 5. **Solve for $r$:** $$r = \frac{14}{\frac{\pi}{2}} = 14 \times \frac{2}{\pi} = \frac{28}{\pi}$$ 6. **Find the central angle for arc $PQ$:** Given as $\frac{2\pi}{3}$ radians. 7. **Find the length of arc $PQ$:** $$\text{Length of arc } PQ = r \times \frac{2\pi}{3} = \frac{28}{\pi} \times \frac{2\pi}{3} = \frac{28 \times 2 \cancel{\pi}}{3 \cancel{\pi}} = \frac{56}{3}$$ 8. **Find the central angle for arc $QR$:** Since $PQR$ is the arc from $P$ to $R$ passing through $Q$, the arc $QR$ corresponds to the central angle: $$\angle QOR = \angle POR - \angle POQ = \frac{\pi}{2} - \frac{2\pi}{3} = \frac{3\pi}{6} - \frac{4\pi}{6} = -\frac{\pi}{6}$$ Since angle cannot be negative, we take the positive equivalent by adding $2\pi$ if needed, but here the arc $QR$ is the remaining part of the circle between $Q$ and $R$ going the other way around. Alternatively, since $PQR$ is the arc from $P$ to $R$ passing through $Q$, and $PQ$ is $\frac{2\pi}{3}$, the arc $QR$ is: $$\angle QOR = \angle POR - \angle POQ = \frac{\pi}{2} - \frac{2\pi}{3} = -\frac{\pi}{6}$$ Negative means the arc $QR$ is the other way around, so the positive angle is: $$2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6}$$ 9. **Calculate the length of arc $QR$:** $$\text{Length of arc } QR = r \times \frac{11\pi}{6} = \frac{28}{\pi} \times \frac{11\pi}{6} = \frac{28 \times 11 \cancel{\pi}}{6 \cancel{\pi}} = \frac{308}{6} = \frac{154}{3}$$ **Final answer:** $$\boxed{\frac{154}{3}}$$ The length of arc $QR$ is $\frac{154}{3}$ units.