Due to symmetry on upper side and lower side, points P and Q are at same potentials. Similarly, points S and T are at same potentials. Therefore, the simple circuit can be drawn as shown below :

$${I_2} = {{12} \over {2 + 2 + 2}} = 2A$$

$${I_3} = {{12} \over {4 + 4 + 4}} = 1A$$

$$\therefore$$ $${I_1} = {I_2} + {I_3} = 3A$$

I_PQ = 0 because V_P = V_Q

Potential drop (from left to right) across each resistance is

$${{12} \over 3} = 4V$$

$$\therefore$$ $${V_{MS}} = 2 \times 4 = 8V$$

$${V_{NQ}} = 1 \times 4 = 4V$$

or, $${V_S} < {V_Q}$$