Final answer:
When the volume of a gas-phase reaction at equilibrium is reduced, the equilibrium will shift towards the side with fewer moles of gas to counteract the increased pressure, provided there is a different number of moles of gases involved. So options (3) is correct.
Step-by-step explanation:
In considering the effect of reducing the volume of a reaction mixture at equilibrium, it's important to apply Le Chatelier's principle which predicts how the system will react to a change in conditions. If reducing the volume causes an increase in pressure, the equilibrium will shift to the side with fewer moles of gas. In the case where the reaction involves different numbers of moles of gases on each side, as in your example where the reactants involve three moles and the products two moles, reducing the volume will cause the equilibrium to shift to the right, toward the products, to counteract the change in volume and pressure. Conversely, if the molar amounts are equal on both sides, this shift will not occur.