This question is not a 3sum problem. The question can be solved with hash tables. For each element in C traverse A and see if the corresponding element exists in B. Time complexity would be $O(|C|*|A|*1) = O(|C||A|)$
Answer to the updated question:
This problem can be treated as 2SUM problem for each element in C (or any other list as they are symmetrical). That is for each $c \in C$ we find solutions for the sum $k - c$ using elements from A and B.
For a $O(1)$ auxiliary space algorithm one can first sort A and B. Then use 2-pointer algorithm (Tutorial) to solve the 2SUM problem in $O(n)$.
Comment if you have doubts on implementation. You can also see https://cs.stackexchange.com/a/13586/81272