**Answer C**

$L_1= \left \{ a^m b^mca^nb^n \mid m,n \geq 0 \right \}$

It is a CFL. It will generate strings which start with a's followed by equal number of b's and single c followed by a's followed by equal number of b's

i.e., ${abcab, aabbcab, aaabbbcaabaabb, \ldots}$

$L_2=\left \{ a^i b^j c^k \mid i,j,k \geq 0 \right \}$

It is a Regular (and of course CFL). It will generate strings which begin with a's followed by any number of b's followed by any number of c's

So, $L_1 \cap L_2$ will be **all strings which are common in both languages**

So, those will be the strings which begin with a's followed by equal number of b's and ending with c

They would be of the form $a^mb^mc$

Such a language is not regular as it requires equal number of a's and b's. But it is Context Free language as we can make a PDA for it. Also, CFLs are closed under intersection with regular set. So, any CFL intersection a regular set gives a CFL (may or may not be a regular set).