A,b and c form a sequence with common difference 150. a = 3p^2q b = 3^2p^3 c = 3p^2r?

a, b and c have a common factor of 3p^2, and are also all multiples of 150. As the other bits of a, b and c are all different, 3p^2 must either be equal to 150, or one of its factors. 150=2x3x5x5=2x3x5^2, so the only possible value for 3p^2 is 75 with p=5. b=9p^3=9x125=1125, therefore a=1125-150=975 and c=1125+150=1275, then q=975/3p^2 = 13 and finally r=1275/3p^2 = 17.