Find out reflexive, symmetric and transitive closure of following relation R. R = {(1,2), (2,3), (3,3)}

Let A={1,2,3}

For R to be reflexive, then it aRa for all a in A. But, $2\not R~ 2$. Hence R is not reflexive.

For R to be symmetric, then $aRb\implies bRa \forall a,b\in A$. But $1R2 \text{ but } 2\not R 1$Hence R is not symmetric.

For R to be transitive, then $aRb \text{ and } bRc, \text{ then }aRc.$ But, 1R2 and 2R3 and $1\not R 3$. Hence R is not transitive.