7805 efficent circuit?
This circuit is no more efficient than just using the 7805 by itself. The 7805 uses BJTs just like you have here to "burn off" excess voltage. The efficiency of linear regulator is really dependent on the voltage that they have to reduce. Standard linear regulators use a Darlington pair of NPN transistors that requires at least 1.5v above the regulated voltage to operate. LDO or low drop out regulators use a single PNP and require about 0.7v above regulated voltage. At just above the operating voltage is where maximum efficiency is achieved because the minimum amount of voltage needs "burned off." 

A basic power calculation is voltage drop time current. 

In your case 12v down to 5v at 3A would give (12 - 5) * 3 = 21W. Your transistor would have to dissipate 21 watts.