Just did some quick calculations...
the diameter of the PL-12 is 0.203m and overall diameter+wingspan is 0.67m. If they are emplaced roughly like so:
Then the total distance width required will be 3.086 meters.
Here's my calculations:
Actual wingspan (i.e.: width of two wings ) = "wingspan" - diameter of missile = (0.67m - 0.203m)
Divide by two to find width of one of the actual "wings", i.e.: the distance "X" which is the distance that will be between each of the missiles (without clearance)
That gives (0.67-0.203)/2 = 0.2335 meters
then add up all the distances.
4 x "distance X" + 3 x "diameter of missile" = length (or width, I suppose) needed to hold three PL-12s.
4 x 0.2335 + 0.203 x 3 = 1.543 meters
Times that by two = total width for six missiles necessary
2 x 1.543 = 3.086 meters
Now assuming there would need to be some clearance between the PL-12s and the "walls" of the weapon bay and between the PL-12s themselves... let's pretend the clearance distance necessary is five centimeters (0.05 meters), then we get...
previous "total distance" + 8 x "clearance distance" = total distance + clearance required
3.086 + 8 x 0.05 = 3.486
Now the F-22's weapon bay width is about 3.9 meters (?) but I'm not sure if that's the actual width or the width with the levers and doors in addition.
Either way, seeing as J-20 has similar if not slightly larger dimensions compared to the F-22 I think six PL-12's in the ventral bay should be quite possible.
I winged this quick type up btw, so I might have overlooked some obvious arithmetic, if so please make your own additions and give an opinion? If anyone can estimate a width of the J-20's underside fuselage that would be helpful too.
Cheers.
(All values taken from wikipedia)