The given Wheatstone bridge is a balanced one.
In the given circuit, the ratio of resistances in the arms of the bridge is

$\frac{P}{Q} = \frac{2}{3} \ldots (i)$

$\frac{R}{S} = \frac{4}{6} = \frac{2}{3} .... (ii)$

Since, $\frac{P}{Q} = \frac{R}{S} = \frac{2}{3}$ , hence the bridge is a balanced one.
Hence, $7\Omega$ resistance is avoided. The given circuit can be reduced as follows :
The $2\Omega$ and $3\Omega$ resistances are connected in series, hence circuit reduces to.
The $5\Omega$ and $10\Omega$ resistances are in parallel

$\frac{1}{R'} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10} \\ \Rightarrow R' = \frac{10}{3} \Omega$