Two airliners, A and B, fly from point P to points Q and R respectively. Given their coordinates and speed, determine which airliner arrives earlier and by how much.

The shortest route between two points on Earth is along a great circle. The angle $\theta$ between two points $(L_1, \lambda_1)$ and $(L_2, \lambda_2)$ on a great circle is given by:

For Airliner A, $P(60^\circ N, 35^\circ W)$ and $Q(75^\circ N, 145^\circ E)$.
$L_1 = 60^\circ$, $L_2 = 75^\circ$.
$\lambda_1 = -35^\circ$ (West), $\lambda_2 = 145^\circ$ (East).
The difference in longitude is $\Delta \lambda = \lambda_2 - \lambda_1 = 145^\circ - (-35^\circ) = 145^\circ + 35^\circ = 180^\circ$.

So, $\theta = \arccos\left(\frac{\sqrt{2}}{2}\right) = 45^\circ$.
Convert $\theta$ to radians: $\theta = 45^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{4}$ radians.
The distance $D_A = R_E \theta$.

Time taken by Airliner A: $T_A = \frac{D_A}{\text{Speed}}$.

Airliner B flies from $P(60^\circ N, 35^\circ W)$ to $R(60^\circ N, 25^\circ E)$ along the parallel of latitude $60^\circ N$.
The distance along a parallel of latitude is given by $D = R_E \cos L \Delta \lambda$, where $\Delta \lambda$ is the difference in longitude in radians.
Here, $L = 60^\circ$.
$\lambda_P = 35^\circ W$, $\lambda_R = 25^\circ E$.
The difference in longitude $\Delta \lambda = 35^\circ + 25^\circ = 60^\circ$.
Convert $\Delta \lambda$ to radians: $\Delta \lambda = 60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3}$ radians.

Time taken by Airliner B: $T_B = \frac{D_B}{\text{Speed}}$.

$T_A \approx 8.338 \, \text{hours}$
$T_B \approx 5.563 \, \text{hours}$
Airliner B arrives earlier.

Difference in time $= T_A - T_B = 8.338 - 5.563 = 2.775 \, \text{hours}$.
Convert to minutes: $2.775 \, \text{hours} \times 60 \, \text{minutes/hour} = 166.5 \, \text{minutes}$.
Rounding to the nearest minute, the difference is $167 \, \text{minutes}$.

Airliner B arrives earlier by approximately 167 minutes.

The final answer is $\text{Airliner B arrives earlier by 167 minutes}$.