As you might have guessed the last language is not the intersection. The first language, as you said, is "all odd strings of length greater than or equal to 3 that end and begin with the same symbol". The second language is "all strings in which a certain symbol appears at the beginning, at the end and at least once anywhere between the two". So for the intersection you need to combine those two propositions with a logical AND and the resulting language would be something like "all odd strings of length >=3 that end and begin with the same symbol and have that same symbol somewhere else inside the string". The tricky part is that symbol Y represents different things in each of the languages. L3 puts an additional constraint saying that the character at the beginning and at the end of the string needs to be exactly at the center too, which is not required by the actual intersection.