Three capacitors C 1 = 2   μ F ,   C 2 = 6   μ F and C 3 = 12   μ F are…

From the above circuit diagram, the equivalent capacitance of the combination of capacitors is, 

$C_{\mathrm{eq}}=C_1+\frac{C_2{C}_3}{C_2{C}_3}$, 

$C_{\mathrm{eq}}=2+\frac{6\times 12}{6+12}=6 \mu \mathrm{F}$

Then, $\frac{Q_{C_1}}{Q_{C_2}}=\frac{C_{1}V}{\frac{C_2{C}_3}{C_2+C_3}V}=\frac{2\times (6+12)}{6\times 12}=\frac{1}{2}$ 

and $Q_{{\mathrm{C}}_2}=Q_{{\mathrm{C}}_3}$

So, the ratio of charges are, $1:2:2$.