The 12 V 3 A supply sounds appropriate, but the 5 V supply is not. If you want to maximize the holding torque, you have to allow the maximum current thru the windings. This can't be achieved at only 5 V according to the specs you quoted.
Added:
Now that you have supplied a link to the datasheet, we know the specs for sure. Your motor is rated for 1 A, 3.8 Ω DC resistance, 3.8 mH inductance, and 75 V max. I have no idea where you got the 12 V ratings from you mentioned in your question.
Since the maximum current is 1 A and the coil has a resistance of 3.8 Ω, you can only apply 3.8 V to a winding continuously. The 75 V rating is the maximum you must not exceed under any circumstances.
Since the motor coils have significant inductance, it takes time to build up current when a fixed voltage is applied. However, torque is proportional to current. This is why stepper motors are often driven with a roughly current controlled supply. The voltage will be high initially as the current in the inductor builds up, then drop to the sustaining level. If you know you are stepping the motor fast, then you can use a higher fixed voltage keeping in mind the time it takes for the current to build to the maximum level given the coil inductance and resistance. Since these parameters are reasonably well known, you can do this open loop. The usual way nowadays is to have a micro do the calculations and control the coil drive with PWM from a fixed high voltage. The duty cycle is changed such that the coil sees a apparent high voltage right after switching, which then decays down to the sustaining voltage as the current builds up. You can also use the same PWM technique but with some current feedback. This feedback is used to adjust the PWM duty cycle to whatever it takes to maintain the desired current.
Before calculations you need to be clear about some fundamental concepts about motors.
"It said that current required by motor = 3 × current required while running on starting."
This statement can not be true, because motor draws more current at start up than it rotates in operational angular velocity.
"When we run the motor on battery eventually battery voltage got dropped, taking more current."
This statement is also false. Motor speed ----> Voltage, Motor torque -----> current.
My advice is as follows. Your application does not require detailed calculations with motor specs. What you need to do is measure the current when your motor rotates under the load, which you choose. Then compare it with battery capacity.
About voltage drop of battery, it does not draw more current when battery gets low on Voltage. Voltage is about speed(Theoretically). Thus, as long as your load is the same, theoretically, motor draws same current.
NOTE: Of course current change when the Armature voltage changes, but this is negligible.
Best Answer
A well designed 12V BLDC fan ought to be able to run 6~16V. This is the absolute max for the Cap voltage if you do not exceed the current spec then conduction losses and self heating will not increase and motor increases efficiency.
but then bearings wear out faster.
for ideal cooling, force turbulent flow at max surface velocity over hotspot with vent design, but use duct for laminar flow of heat removal and avoid grid block at blade surface for eddy current noise and flow restriction.