`3sin^2A+2sin^2B = 1`
`=>3sin^2A = 1-2sin^2B`
`=>3sin^2A = cos2B->(1)`
Now,
`3sin2A-2sin2B = 0`
`=>3sin2A = 2sin2B`
`=>sin2B = 3/2sin2A->(2)`
Now, `cos(A+2B) = cosAcos2B-sinAsin2B`
`=3cosAsin^2A-3/2sinAsin2A`
`=3cosAsin^2A-3/2sinA(2sinAcosA)`
`=3cosAsin^2A-3cosAsin^2A`
`= 0`
`:. cos(A+2B) = 0`
`=>A+2B = pi/2`
so, option `(b)` is the correct option.