Capacitors 1 and 2 are in parallel, so their capacitance adds up to $10\mu F + 10\mu F = 20\mu F$. This combination is in series with capacitor 3, so the equivalent capacitance of this part of the circuit is $\frac{1}{\frac{1}{20} + \frac{1}{10}} = 6.67\mu F$.
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