Since line segments \( AB \) and \( XY \) are congruent and parallel, translating them will maintain their parallelism and the distance between them.
Given that segments \( AB \) and \( XY \) are translated 2 units left and 3 units down, the translation affects their positions but not the distance between them. The segments will move but will remain parallel and congruent to each other.
If the original distance between the segments \( AB \) and \( XY \) is \( d \), after the translation \( A'B' \) and \( X'Y' \) will remain the same distance apart. Since the translation does not affect the distance between parallel lines, the distance remains \( d \).
To find the distance between \( A'B' \) and \( X'Y' \):
- The translation is 3 units down, and that shifts both segments downward.
- Since \( AB \) and \( XY \) are originally parallel and congruent, the distance between them before the translation remains the same after the translation.
Therefore, the distance between line segments \( A'B' \) and \( X'Y' \) is still \( d \).
If there are numerical values given for the original distance \( d \), that value would be the final answer. In a general case, we can let \( D \) be the distance, which is unchanged by the translation.
So the answer is that the distance between \( A'B' \) and \( X'Y' \) is the same as the distance between \( AB \) and \( XY \).
Let’s denote the distance between the original segments as \( D \). Therefore, the distance between the translated segments is:
\[ \text{Distance between } A'B' \text{ and } X'Y' = D. \]
Without specific numerical values, we conclude based on the problem's conditions that the distance remains the same.