Three resistances P, Q, R each of 2 and an unknown resistance S...

The situation can be depicted as shown in figure. As resistances S and $6 \Omega$ are in parallel their effective resistance is $\frac{6 S}{6+S} \Omega$ As the bridge is balanced, hence it is balanced Wheatstone's bridge. For balancing condition, $$\begin{aligned} & & \frac{P}{Q} & =\frac{R}{\left(\frac{6 S}{6+S}\right)} \\ \text{ or } & & \frac{2}{2} & =\frac{2(6+S)}{6 S} \\ \text{ or } & & 3 S & =6+S \\ \text{ or } & & S & =3 \Omega \end{aligned}$$