In the circuit diagram shown in figure which option is incorrect

Capacitors 1$\mathrm{\mu }$ F,2$\mathrm{\mu }$ F and 3$\mathrm{\mu }$F are in parallel, and their total capacitance is 6$\mathrm{\mu }$F .
Thus, we have three capacitors in series, each of capacitance 6$\mathrm{\mu }$ F across the 12 V
power supply. So potential drop across each is 12 / 3 $\mathrm{\mu }$ 4V . So options (3) is correct.
Charge on 2$\mathrm{\mu }$ F capacitor is CV = 2$\mathrm{\mu }$ 4 $\mathrm{\mu }$ 8$\mathrm{\mu }$C. So option (1) is correct.
Charge on 6$\mathrm{\mu }$ F capacitor $6\times {10}^{-6}\times 4=24\mu C$ . So option (2) is wrong.
Thus, correct options are (1), (3), and (4) and option (2) is correct.