The distance between the places P and Q is D units. The average speed that gets a person from P to Q in a stipulated time is S units. He takes 20 minutes more time than usual if he travels at 60 Km/h, and reaches 44 minutes early if he travels at 75 Km/h. The sum of the thrice the numerical values of D and thrice the numerical value of S is:

Let usual time be t hours, stipulated time. Distance D = S × t. At 60 km/h: time = t + 20/60, so D = 60(t + 1/3) = 60t + 20. At 75 km/h: time = t - 44/60, so D = 75(t - 11/15) = 75t - 55. Equating: 60t + 20 = 75t - 55 → 15t = 75 → t = 5 hours. Then D = S × 5, and also D = 60 × 5 + 20 = 320, so S = 64 km/h. 3D + 3S = 3(320) + 3(64) = 960 + 192 = 1152. The problem involves two scenarios with different speeds and time deviations from stipulated time.