Limiting the current on a PC power supply
The Power Supply Unit (PSU) creates a voltage difference on its leads.  This means that the 12 Volts output connector  will have a 12 Volt difference between its leads. Note that this says nothing about current yet.

When you connect the 12V leads to a load (think of a load as a resistor), the current through the load will always follow Ohms Law: \$I_{}\$ = \$V_{}\$/\$R_{}\$. In our case, the load is your motor.

The rating on your PSU is the upper limit that the PSU is telling you it will handle.  If your load (\$R_{}\$) is so small that the current draw (\$I_{}\$) is higher than 34A, the PSU will fail in one way or another. With your three-motor system, the maximum draw that the power supply will see is 9A. This is well below the PSU's maximum output capability (34A). 

The 3A rating on your motors is the amount of current that they will effectively draw from a 12VDC supply. You can thus deduce that each motor's effective on-resistance is:

\$R_{}\$=\$V_{}\$/\$I_{}\$ 

\$R_{}\$ = 12/3 = 4 Ohms

With the 4 Ohm load (your motor), your supply will pump only 3A per motor. If you want to prevent any accidental draw of over 3A, you would install an in-line safety fuse for each of the motors. The fuse will self-destruct (and force an open circuit) if more than 3A is pumped through it, saving your motors. 

Hope this helps!