Calculate the total distance between Town A (60°N, 40°W) and Town C (10°N, 140°E), passing through Town B (60°N, 140°E), using the Earth's radius R = 6371 km.

Town A is at $(60^\circ N, 40^\circ W)$ and Town B is at $(60^\circ N, 140^\circ E)$.
Since both towns are on the same latitude ($60^\circ N$), the plane flies along a parallel of latitude.
The difference in longitude $\theta$ is $40^\circ W + 140^\circ E = 180^\circ$.
The formula for distance along a parallel of latitude is $D = \frac{\theta}{360^\circ} \times 2\pi R \cos\alpha$, where $\alpha$ is the latitude.

$$D_{AB} = \frac{180^\circ}{360^\circ} \times 2\pi (6371 \text{ km}) \cos(60^\circ)$$ $$D_{AB} = \frac{1}{2} \times 2\pi (6371 \text{ km}) \times 0.5$$ $$D_{AB} = \pi (6371 \text{ km}) \times 0.5$$ $$D_{AB} = 3185.5\pi \text{ km}$$ $$D_{AB} \approx 10007.54 \text{ km}$$

Town B is at $(60^\circ N, 140^\circ E)$ and Town C is at $(10^\circ N, 140^\circ E)$.
Since both towns are on the same longitude ($140^\circ E$), the plane flies along a meridian (a great circle).
The difference in latitude $\phi$ is $60^\circ N - 10^\circ N = 50^\circ$.
The formula for distance along a great circle (meridian) is $D = \frac{\phi}{360^\circ} \times 2\pi R$.

$$D_{BC} = \frac{50^\circ}{360^\circ} \times 2\pi (6371 \text{ km})$$ $$D_{BC} = \frac{5}{36} \times 2\pi (6371 \text{ km})$$ $$D_{BC} = \frac{5}{18}\pi (6371 \text{ km})$$ $$D_{BC} \approx 5554.08 \text{ km}$$

The total distance covered by the plane is the sum of $D_{AB}$ and $D_{BC}$.

$$\text{Total Distance} = D_{AB} + D_{BC}$$ $$\text{Total Distance} \approx 10007.54 \text{ km} + 5554.08 \text{ km}$$ $$\text{Total Distance} \approx 15561.62 \text{ km}$$

$$\text{Total Distance} \approx 15562 \text{ km}$$

The total distance covered by the plane is $\text{15562 km}$.