`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.