Three capacitors C _ 1 = 2 , C _ 2 = 6 and C _ 3 = 12 are connected...

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

${C}_{eq} = {C}_{1} + \dfrac{{C}_{2}{C}_{3}}{{C}_{2}{C}_{3}}$,

${C}_{eq} = 2 + \dfrac{6 \times 12}{6 + 12} = 6\muF$

Then, $\dfrac{{Q}_{{C}_{1}}}{{Q}_{{C}_{2}}} = \dfrac{{C}_{1}V}{\dfrac{{C}_{2}{C}_{3}}{{C}_{2} + {C}_{3}}V} = \dfrac{2 \times ( 6 + 12 )}{6 \times 12} = \dfrac{1}{2}$

and ${Q}_{{C}_{2}} = {Q}_{{C}_{3}}$

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