That is not the right topology for using transistors to increase the current of a linear regulator. Here is how it's done using a single transistor to provide more current:
This still keeps the output voltage well regulated. In your circuit, the B-E drop of the transistors will make the output voltage lower.
At low currents, there is little voltage accross R1, so Q1 stays off. When the load current increases, the voltage accross R1 increases, which turns on Q1, which dumps more current onto the output. The regulator is still regulating, but the current thru it will stop increasing at around 3/4 Amp in this case, after which the transistor takes over most of the additional load.
One big power transistor with a big heat sink should be able to handle your 10 A output current. However, if you want to spread the heat accross multiple transistors, you can't just add more of them in parallel. The way to add more transistors is to give each its own emitter resistor. This provides a little negative feedback so that if a transistor is passing more than its share of the current, the voltage accross its emitter resistor will be higher, which will take away from its B-E voltage, which will decrease the current thru the resistor.
Here is a example with 3 external transistors that take most of the current load, while the regular is providing the regulation:
This is basically the same idea as before, but each transistor has its own emitter resistor. R1 is also increased a little bit to make sure there is plenty of base drive available for all three transistors, and to account for additional voltage drop accross the emitter resistors. Still, R1 is larger than it needs to be in this example. However, you have plenty of headroom voltage available, so dropping a little more in a resistor is no problem.
Keep the dissipation of the resistors in mind. Let's say to account for a little imbalance and some margin, we want each of the transistor to be able to handle 4 A. That is 400 mV accross the emitter resistor, plus 750 mV or so for the B-E drop, for a total of 1.15 V that needs to be accross R1 at full current. That means it will dissipate 660 mW, so it needs to be at least as "1 W" resistor.
Each emitter resistor must be able to safely dissipate (4 A)2(100 mΩ) = 1.6 W. These should be at least "2 W" resistors.
All this said, I agree with Wouter in that this is the wrong way to address your overall problem. Linearly regulating down 12 V to make 5 V will be more trouble and a lot more wasteful than a switcher. However, the real way to address this is to step back a few levels and re-think at the system level. Running lots of high current stuff at 5 V from a 12 V battery makes little sense. You should be able to find motors that run at 12 V, actually more easily than ones that run at 5 V at this power level. You then only need to provide 5 V for the control logic, which controls switches that enable power to the 12 V devices. Or you can still use 5 V devices with a proper PWM drive so that you are switching the 12 V on and off fast enough so that the devices only see the average of 5 V.
There should be several good options at the system level, none of which include wasting 70 W as heat to run 5 V motors from 12 V.
I described how to make a higher current linear regulator from a existing one and some external transistor to document how to do it right, but this should not really be part of your overall solution.
Best Answer
Although it can be calculated, here are the modeled (optimized) values with 1% resistors.