As long as the set has, in this case, (1,1), (2,2), and (3,3), then its reflexive - the fact that 1 is ALSO, for example, equivalent to 2 has no bearing on the reflexivity. 1 is related to 1, 2 to 2, 3 to 3, and that's what you need.

For symmetry, you need to show that if a is related to b then b is related to a for ALL a and b.

Here, you don't need to worry about (1,1), (2,2), and (3,3) - clearly they're symmetric. Consider (1,2) - this means 1 is related to 2. For R to be symmetric, 2 must also be related to 1, i.e., (2,1) must be in your set (which it is). That takes care of everything - you've checked and "paired up" every element in R - note that (1,1), (2,2), and (3,3) are their own "pairs."

For transitivity, you need to show that if (a,b) is in R and (b,c) is in R, then (a,c) is in R. (In other words, if a is related to b and b is related to c, then a is related to c.)

Consider (1,2) in combination with (2,1) for example. We have:

1 is related to 2.

2 is related to 1.

The question for transitivity, then, is: "Is 1 related to 1"? It is, since (1,1) is also in the set.

Make sense?

Hope this helps!