Correct Answer - `6.49 muC`, positive
as shown in Fig, force exertend on charege `+3muC` by charge at B
`F_(1) = (1)/(4pi in_(0)) (q_(1) q_(2))/(r^(2)) = (9xx10^(9)xx3xx10^(-6)xx5xx10^(-6))/((0.2)^(2))`
`= 3.375 N` along AB
Force exerted on `+3 muC` charge by charge at C
`F_(2) = (9xx10^(9)xx3xx10^(-6)xx5xx10^(-6))/((0.2)^(2))`
`= 3.375 N` along Ac
Resultant force of `F_(1) and F_(2)`
`F = sqrt( F_(1)^(2) + F_(2)^(2) + 2 F_(1) F_(2) cos 60^(@)) = 5.845 N`
For charge A to be in equlibrium positive charge must be placed at M, to exert force along MA
`AM = sqrt((0.2)^(2) - (0.1)^(2)) = 0.1xx sqrt(3) m`
Force on charge `+3 muC` will be zero if
`(9xx10^(9) xx q xx3xx10^(-6))/((0.1xx sqrt(3))^(2)) = 5.845`
`q = 6.49 muC`