As mentioned above, using four 3.3V LEDs in series with a 12V supply isn't a great idea.
For starters, you won't be running the LEDs at peak brightness as the maximum voltage drop across each one will only be 3V, not 3.3V (although to be fiar you probably won't even notice this).
A more serious issue is that because your current limiting resistor is so small, the power dissipated in it becomes massively sensitive to the voltage dropped across the LEDs.
With that amount of LEDs you will get one or two fail (my home made lounge lights use 160 LEDs and I regularly get failures). LEDs can fail open or closed circuit. If one fails open circuit then it behaves like a fuse and protects the rest of the circuit - all well and good. If it fails closed then you have to make sure that the increased current flow does not cause any secondary damage.
See the bit at the bottom of your calculator where it says that "each 1 ohm resitor dissipates 0.4mW" and "the wizard thinks that 1/4W resistors are fine for your application"? Well, if all LEDs are working then that's true, but if you have one LED fail closed-circuit then the current in that part of the circuit will rocket up to about 3A. This will put 9W into your current limiting resistor. If the resistor or one of the other LEDs doesn't blow open-circuit, and if you don't have a fuse somewhere then you could get all sorts of nasty things happening, including all the magic smoke coming out

. 3A is a LOT of current.
The maths in the online calculator is correct, but unfortunately its not good circuit design.
Using 3 LEDs in series and 150ohm resistors would enable you to be safe even if you had one LED failing closed circuit in each branch of the array, even with 1/4W resistors (but I would use 1/2W to be extra safe.