*L*_{1 = Regular because we can give a REGEX: 0+1+}

*L*_{2 = 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.*

*L*_{3 }= {0^{p}1^{p}0^{p }| 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*