L1 = Regular because we can give a REGEX: 0+1+
L2 = equal number of 0's and equal number of 1's and after a "1" you can never see a "0"
Take a valid string from L2 = "0011"
Push all "0's"and for every "1" pop one "0" Hence it's DCFL because we know when and what to Push and Pop and every DCFL is CFL.
L3 = {0p1p0p | p>0, p=q=r} and this is classic CSL.
p>0 because we have given p=q=r therefore I can write it as "p" in place of q and r and since p,q,r belongs to N (Natural Number) which makes p>0
Hence Answer is c) because L1 = L2 = CFL, but L3 = CSL, not CFL