Let’s break this down into 2 cases:
A = Each even position has 3 choices: $ a, b, c $
There are 6 even positions, so there are $ 3^6 $ ways to choose the characters for the even positions.
$ |A| = 3^6 $
B = Each odd position has 2 choices: $ d, e $
There are 6 odd positions, so there are $ 2^6 $ ways to choose the characters for the odd positions.
$ |B| = 2^6 $
$ |A \cap B| = 3^6 \cdot 2^6 $
$ |A \cap B| = 3^{12} $
$ |A \cap B| = 6^6 $