1. Given relation

On the set $A=\{3,6,9,12\}$ we have $R=\{(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)\}.$

We check whether $R$ is reflexive, symmetric, and transitive.

2. Check reflexive

A relation on $A$ is reflexive if for every $a\in A$, we have $(a,a)\in R$.

Elements of $A$ are $3,6,9,12$. We see that

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

So $R$ is reflexive.

3. Check symmetric

A relation is symmetric if whenever $(a,b)\in R$, then $(b,a)\in R$ also.

Now,

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

Therefore, $R$ is not symmetric.

So options A and B are immediately false.

4. Check transitive

A relation is transitive if whenever $(a,b)\in R$ and $(b,c)\in R$, then $(a,c)\in R$.

We verify all nontrivial chains.

From element $3$

We have $(3,3),(3,6),(3,9),(3,12)\in R.$

• With $(3,3)$ followed by any of $(3,3),(3,6),(3,9),(3,12)$, we need $(3,3),(3,6),(3,9),(3,12)$ respectively — all are in $R$.
• $(3,6)$ and $(6,6)$ imply $(3,6)\in R$.
• $(3,6)$ and $(6,12)$ imply $(3,12)\in R$.
• $(3,9)$ and $(9,9)$ imply $(3,9)\in R$.
• $(3,12)$ and $(12,12)$ imply $(3,12)\in R$.

So all chains starting from $3$ satisfy transitivity.

From element $6$

We have $(6,6),(6,12)\in R.$

• $(6,6)$ and $(6,6)$ imply $(6,6)\in R$.
• $(6,6)$ and $(6,12)$ imply $(6,12)\in R$.
• $(6,12)$ and $(12,12)$ imply $(6,12)\in R$.

All required pairs are present.

From element $9$

We have only $(9,9)\in R.$ Then $(9,9)$ and $(9,9)$ imply $(9,9)\in R$, which is true.

From element $12$

We have only $(12,12)\in R.$ Then $(12,12)$ and $(12,12)$ imply $(12,12)\in R$, which is true.

Hence $R$ is transitive.

5. Conclusion

The relation is:

• reflexive: Yes
• symmetric: No
• transitive: Yes

Therefore, the correct option is $\boxed{\text{D: reflexive and transitive only}}.$

6. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.