maybe i'm a little dull here... but i'm going to take a shot at this.
I think...
VBEQ1 is the voltage drop between the base and emitter of Q1.
BQ1 is the DC current Gain (Or hfe) of the transistor.
Looks like your current summing the outputs... using the LM78xx to get a fixed voltage, which Q1 will also track, and then driving more current. How much I'm not sure of.
The 7818's can handle 1A apparently (That would be IREG). The current output of transistor Q1 is given by BQ1*(IREG - VBEQ1/R1)
So, if we throw in the values
IREG = 1A
BQ1 = 25 (datasheet)
VBEQ1 = 1.5V (maximum)
R1=3Ohm
Then we get
IOUT = 1 + 25(1-(1.5/3))
= 13.5A
However, it's best we get some real analog designers to verify that. Also, your still dropping volts accross Q1 - which is gonna make it a little warm.