This thread is getting increasingly silly. 'theelectronicsfan' clearly wants to ignore the fundamentals rules of electronics in the hope it will save some time to unsolder all the LEDs.
I'll give it one more try:
If you join them in series, that means in a chain with the + end of one diode connected to the - of the next in the chain, they will all pass the same current and be roughly the same brightness. The voltage you need to operate them is a little higher that the total of all the LED voltages added together. You need one resistor to limit the current for the whole chain and it's value is (Supply voltage - total LED voltage) / LED current. The power the resistor will dissipate is (LED current * LED current) * resistor value.
In parallel, as stated many times already, the diode with the lowest voltage drop will decide the voltage across all the other diodes, it will 'hog' the current at the expense of all the others. All the LEDs have the same voltage across them (they are wired directly across each other so that must be the case) so suppose the LED with the lowest voltage is 2V and another is 2.2V, the voltage will be held at 2V by the the lowest voltage LED and the other one will be dim at best, it doesn't have enough voltage to fully light up. If you increase the current by dropping the resistor, the voltage will eventually rise and the dimmer LEDs will get brighter, but only because the lowest voltage LEDs are now burning out.
You have two options to fix your problem, playing with one resistor value is never going to work, whatever its value and whatever it's power rating.
Option 1. (this is best)
Split your LEDs into several smaller series chains. You are starting with 12V and your LEDs have a maximum forward drop of 2.2V so you can safely wire 5 LEDs in one chain. That gives a maximum requirement of 5 x 2.2 = 11V. In each chain, add a resistor of (12-11)/0.02 = 50Ω. It will dissipate (1*1)*0.01 Watts so even a tiny resistor will do. Wire each of the chains in parallel.
Option 2.
Wire all the - ends of the LEDs together and connect the back to the power source, add a resistor in the + end of every diode and connect the resistor back to the power source. The resistor needs to be (12-2.2)/0.02 = 490Ω and it will dissipate 0.2W, still quite small.
Option 1 will only need one resistor per 5 LEDs, option 2 needs the same number of resistors as LEDs and produces more heat. The choice is yours but I (an all the other contributors) assure you there is no magical way of making your present circuit work as you intend. You will have to rewire it.
Note that I have used 20mA (0.02A) in my calculations above.
Brian.