Three resistances of 1Ω, 2Ω and 3Ω are given. In order to obtain an equivalent resistance of $\frac{11}{3}Ω$ they are to be connected as

3Ω in series with parallel combination of 1Ω and 2Ω

The correct answer is Option (1) → 3Ω in series with parallel combination of 1Ω and 2Ω

Given resistances: $R_1 = 1\,\Omega$, $R_2 = 2\,\Omega$, $R_3 = 3\,\Omega$

Check option 1: 3Ω in series with parallel of 1Ω and 2Ω

Parallel: $R_p = \frac{1 \cdot 2}{1+2} = \frac{2}{3}\,\Omega$

Series with 3Ω: $R_{\text{eq}} = 3 + \frac{2}{3} = \frac{11}{3}\,\Omega$ ✅

Answer: 3Ω in series with parallel combination of 1Ω and 2Ω