x,12,y in H.P.
So, 12=x+y2xy⇒12(x+y)=2xy.....(1) x,12,z,y are in increasing A.P.
So, 24=x+z (option B)...(2) 2z=12+y.....(3) 12+z=x+y.....(4)
From (2), (3) and (4) put values in (1) 12(12+z)=2(24−z)(2z−12)
we get, z=15
Put z=15 in equation (2) and (3), we get x=9,y=18
So, maximum value of (x−3)sinα−(y−10)cosβ+2=6sinα−8cosβ+2
Maximum when sinα=1 and cosβ=−1 is 4 .