A circuit consists of three capacitors C_1 = 2\ , C_2 = 3\ , and...

The capacitors $C_1, C_2$, and $C_3$ are connected in parallel. The voltage across this parallel combination is the same for each capacitor.
Voltage across the parallel group, $V_p = \dfrac{Q_1}{C_1} = \dfrac{12\ \mu\text{C}}{2\ \mu\text{F}} = 6\text{ V}$.

The equivalent capacitance of the parallel group is:
$C_p = C_1 + C_2 + C_3 = 2\ \mu\text{F} + 3\ \mu\text{F} + 5\ \mu\text{F} = 10\ \mu\text{F}$.

The total charge on the parallel group is:
$Q_p = C_p \times V_p = 10\ \mu\text{F} \times 6\text{ V} = 60\ \mu\text{C}$.

Since capacitor $C_4$ is connected in series with the parallel group, it carries the same total charge as the parallel group.
Therefore, the charge on $C_4$ is $Q_4 = 60\ \mu\text{C}$.

The voltage across $C_4$ is:
$V_4 = \dfrac{Q_4}{C_4} = \dfrac{60\ \mu\text{C}}{10\ \mu\text{F}} = 6\text{ V}$.

The total voltage of the source $V$ is the sum of the voltages across the parallel group and $C_4$:
$V = V_p + V_4 = 6\text{ V} + 6\text{ V} = 12\text{ V}$.

Answer: $12\text{ V}$ and $60\ \mu\text{C}$