Description for Correct answer:
Since \( \Large \left(1,\ 1\right);\ \left(2,\ 2\right);\ \left(3,\ 3\right) \epsilon\ R \), therefore R is reflexive.

\( \Large \left(1,\ 2\right)\ \epsilon\ R\ but\ \left(2,\ 1\right) \notin \ R \) , therefore R is not symmetric. It can be easily seen that R is transitive.