What would be the input current of buck regulator
It's closer to your second option, but not quite. 

A buck regulator will take a few mA for its operation, which could be considered to be constant. Say 15mA at 12V. Let's ignore that for the moment. 

If perfect, it would output the same power as input. Your output of 3.3V at 2A is 6.6W, so a perfect buck regulator would need 550mA from the 12V supply (6.6W/12V). (Image from the Wikipedia page) 



To see this, consider that a perfect switch has no losses, a perfect diode either conducts like a wire (no power is lost) or blocks perfectly, and a perfect inductor with zero resistance and zero eddy current losses dissipates no power (it simply stores energy and barfs it up again). So from conservation of energy, input power (averaged over a cycle in steady state) must equal output power. 

In a real buck regulator implementation, there are, however, losses that are at least roughly proportional to the output power due to resistance in the inductor and switch, voltage drop in the diode and so on. Say the efficiency is 80%. It will then require 6.6W/0.80 = 7.5W input power, or 625mA, plus the 15mA it needs just sitting there, so 640mA total (usually the quiescent current will be folded into the overall efficiency figure so we don't need to add it).  

Because of the 15mA (or whatever it is for your real regulator) the efficiency of the regulator (output power divided by input power) will decrease for lower output currents.