Correct Answer - C
`q_(2)-q_(1)=-2Q qQ_(1)-q_(2)+q_(3)=4Q`
`rArr (q_(1))/(4piepsilon_(0))-(q_(1))/(4piepsilon_(0)(2R))+(q_(2))/(4piepsilon_(0)2R)`
`-(q_(2))/(4piepsilon_(0)4R)+(q_(3))/(4piepsilon_(0)4R)`
` V_(A)(1)/(4piepsilon_(0))((q_(1)-q_(1)+q_(2)-q_(2)+q_(3)))/(4R)=V_(c)`
`V_(A)=V_(C)=q_(1)-(q_(1))/(2)+q_(2)/(2)-q_(2)/(4)+(q_(3))/(4)=(q_(3))/(4)`
`rArr (q_(1))/(2)+(q_(2))/(4)=0 rArr q_(2)=-2q_(1)rArr -2q_(1)-q_(1)=-2Q`
`rArr q_(1)=(2Q)/(3)`
rArr the charge that flows through the switch is `Q//3`.