Note that $(X \cap Y^c \cap Z) \cap (X \cap Y \cap Z^c) = \emptyset$, since $Z \cap Z^c = \emptyset$. Thus, the events $A \cap B \cap C^c$, $A \cap B^c \cap C$ and $A^c \cap B \cap C$ are mutually disjoint. So, we are interested in finding:
$$ P(E) = P(A \cap B \cap C^c) + P(A \cap B^c \cap C) + P(A^c \cap B \cap C)$$
The problem tells us that the three events $A, B, C$ are independent; note that this is different from $A, B, C$ being mutually independent, as we only have:
$$ P(A \cap B \cap C) = P(A)P(B)P(C) $$
The solution to the problem hinges upon the following sub-problem: can we determine $P(A \cap B \cap C^c)$ using the information given?
If yes, then by symmetry, we should be able to determine the probability of $P(A \cap B^c \cap C)$ and $P(A^c \cap B \cap C)$.
I am not able to figure out $P(A \cap B^c \cap C)$ using the given information. Does we not have enough information?