Correct Answer - C
(1) `((150-x))/3+((600-x))/6-200=0`
`x=-100 muC`
Hence, final charge on `6 muF` Capacitor is
`q_(1)=700 muC` and charge on `3 muF` is `q_(2)=250 muC`
`q_(1)=(600-x) q_(2)=(150-x)`
. Plates `2` and `3` and plates `1` and `4` form isolated system.
Hence `q_(1)=q_(2)=q_(3)=x=-100 muC`.