In a balanced Wheatstone’s network, the resistances in the arms Q and S are interchanged. As a result of this :

We know that in In a balanced Wheatstone's network, 

$\frac{P}{Q}=\frac{R}{S}$

If Q is replaced by S then the ratio becomes $\frac{P}{S}\ne \frac{R}{Q}$​ which is not equal to the previous ratio. 

Hence, the network is not balanced.