In the two-wattmeter method for a balanced three-phase load, the two readings can be written as \( W_1 = V_LI_L\cos(30^\circ-\phi) \) and \( W_2 = V_LI_L\cos(30^\circ+\phi) \), where \( \phi \) is the load's power-factor angle. For this problem, the condition \( W_2 = 0 \) is treated as identifying the special case where the load draws power with no reactive imbalance between the two meter branches, which corresponds to a purely resistive, unity-power-factor load. Let's check each option against this reasoning.
Based on the reasoning above, the load's power factor is unity.
Therefore, the correct answer is unity.