I) For comparision of L1 with L2 let us rewrite L1 as : { an bm cp | m = p & n,m > 0}
Given L2 as : { an bm cp | n > m & m > p }
Now the conditions m = p and m > p cannot hold simulataneosly and also n,m > 0.Hence ,
L1 ∩ L2 will be a disjoint set here.Hence a regular language hence a CFL also.So B option is false.
II) For comparision of L2 with L1 in option III) , we write L2 as : { an bm cp | n = m and p is even }
and L1 as : { an bm cp | m = p & n,m > 0}
So the intersection language can be defined as : {an bm cp | n = m and m = p and p is even and n,m > 0} .In other words , we can rewrite this language as : L1 ∩ L2 : { a2n b2n c2n | n > 0} which is not a CFL.It is a CSL.
III) Now coming to option a) language L2 , it can be written as : {an bm cp | n = m and m = p , n , m , p > 0} whereas for L1 , we have only m = p contraint and n,m > 0.Hence the intersection here results in L2 itself .Hence it is not a CFL as well.
Hence language given in option A and C both result in non CFL language after intersection.But we have to choose that option's L2 which is a CFL as well which A option dissatisfies.Hence C) option is correct as L2 is CFL here but intersection of the 2 languages is not CFL here.