Will LD33V voltage regulator overheat with 12V 2.4A dc input?
Check the datasheet and do the math. Short answer: YES, it will overheat A LOT!

With that huge 8.7 V voltage drop, if the load required the full 0.8 A that the LD1117V33 can output then the power dissipation would be approx 7 W, which is well below the 15 W maximum rating.

However, the temperature rise at 7 W dissipation without using a heatsink (50 C/W) would be a mind-boggling 350 C, and the junction temperature would go far above the maximum 150 C rating.

At Tamb = 25 C, the maximum allowable temperature raise would be 125 C. At 50 C/W the maximum allowable power dissipation for that temperature rise would be 2.5 W. This would mean a maximum 0.36 A current draw from the load. And you should derate this to 50% (or 70% at most) for reliability reasons, so let's say your maximum current draw should be 180-250 mA.

If your load draws more current than that, then you'll need a heatsink in order to reduce the temperature rise. The maximum thermal resistance (junction to ambient) for a temperature rise of 125 C at 7 W dissipation is 17.8 C/W. The case itself has 3 C/W (junction to case), so the heatsink should have a maximum of 14.8 C/W (case to ambient). You should derate this too, so you actually want a maximum of 7.4-10.4 C/W for your heatsink.

However, the best solution by far is to use a switching regulator / DC-DC  step down converter. A linear regulator is very inefficient power-wise, when used with such huge voltage drops. The switching regulator will not need a heatsink because it doesn't dissipate that much power to begin with.