Let $$R = \{(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)\}$$ be a relation on the set $$A = \{3, 6, 9, 12\}$$. The relation is

A relation on a set can be tested for three standard properties.

• Reflexive: $$(a,a)\in R$$ for every $$a\in A$$.
• Symmetric: whenever $$(a,b)\in R$$, the pair $$(b,a)$$ must also belong to $$R$$.
• Transitive: whenever $$(a,b)\in R$$ and $$(b,c)\in R$$, the pair $$(a,c)$$ must belong to $$R$$.

The set here is $$A=\{3,6,9,12\}$$ and the relation is
$$R=\{(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)\}.$$

1. Reflexive property
We need $$(3,3),(6,6),(9,9),(12,12)$$. All four ordered pairs are explicitly present in $$R$$, so $$R$$ is reflexive.

2. Symmetric property
Check each non-diagonal pair in $$R$$:

• $$(6,12)\in R$$ but $$(12,6)\notin R$$.
• $$(3,9)\in R$$ but $$(9,3)\notin R$$.

The required reversed pairs are missing, so $$R$$ is not symmetric.

3. Transitive property
List all combinations where the second component of the first pair equals the first component of the second pair and verify the third pair:

$$(3,6)\in R,\;(6,12)\in R\;\Rightarrow\;(3,12)\in R$$ (present).
$$(3,9)\in R,\;(9,9)\in R\;\Rightarrow\;(3,9)\in R$$ (present).
$$(6,12)\in R,\;(12,12)\in R\;\Rightarrow\;(6,12)\in R$$ (present).
All other possible middle elements (3, 6, 9, 12) lead to cases that reduce to already listed diagonal pairs, which are in $$R$$ by reflexivity.

Since every required third pair is present, $$R$$ is transitive.

Conclusion
The relation $$R$$ is reflexive and transitive but not symmetric, hence it is not an equivalence relation.

Option A which is: reflexive and transitive only