In a circuit, three capacitors C_1, C_2 and C_3 are connected in...

Let the junction of $C_1$ and $C_2$ be node $B$, and the junction of $C_2$ and $C_3$ be node $C$.

The network can be redrawn to reveal a Wheatstone bridge configuration between terminals $A$ and $D$.
The arms connected to terminal $A$ are $C_1$ (between $A$ and $B$)$and C_4$ (between $A$ and $C$).
The arms connected to terminal $D$ are $C_5$ (between $B$ and $D$)$and C_3$ (between $C$ and $D$).
The central capacitor is $C_2$ (between $B$ and $C$).

Checking the balance condition:
$\dfrac{C_1}{C_5} = \dfrac{4}{12} = \dfrac{1}{3}$
$\dfrac{C_4}{C_3} = \dfrac{2}{6} = \dfrac{1}{3}$

Since $\dfrac{C_1}{C_5} = \dfrac{C_4}{C_3}$, the bridge is balanced. The potential difference across $C_2$ is zero, so it can be removed from the circuit.

The equivalent circuit consists of two parallel branches:
Branch 1 ($A-B-D$): $C_1$ and $C_5$ in series. $C_{branch1} = \dfrac{4 \times 12}{4 + 12} = \dfrac{48}{16} = 3 \mu\text{F}$
Branch 2 ($A-C-D$): $C_4$ and $C_3$ in series. $C_{branch2} = \dfrac{2 \times 6}{2 + 6} = \dfrac{12}{8} = 1.5 \mu\text{F}$

Total equivalent capacitance $C_{eq} = C_{branch1} + C_{branch2} = 3 + 1.5 = 4.5 \mu\text{F}$.

Answer: $4.5 \mu\text{F}$