Let $A=\{1,2,3\}.$ A relation on $A$ is a subset of $A\times A$, where $|A\times A|=9$.

We need relations $R$ such that:

1. $(1,2)\in R$
2. $R$ is reflexive
3. $R$ is transitive
4. $R$ is not symmetric
5. $R$ has at most 6 elements

1. Reflexive condition

Since $R$ is reflexive, it must contain $(1,1),(2,2),(3,3).$ So 3 elements are already compulsory.

Also given: $(1,2)\in R.$ Thus at least 4 elements are fixed: $\{(1,1),(2,2),(3,3),(1,2)\}.$

Since relation can have at most 6 elements, we may add at most 2 more elements from the remaining 5 pairs: $(1,3),(2,1),(2,3),(3,1),(3,2).$

2. Not symmetric condition

Because $(1,2)\in R$, for symmetry we would need $(2,1)\in R$. So for the relation to be not symmetric, it is enough that $(2,1)\notin R.$ If $(2,1)$ were included, we would still need to check other pairs; but since we want not symmetric and have a small case analysis, we will examine all valid possibilities.

3. Transitivity implications

Start with the compulsory part: $R_0=\{(1,1),(2,2),(3,3),(1,2)\}.$ This is already transitive:

• $(1,1)$ with $(1,2)$ gives $(1,2)$, already present.
• $(1,2)$ with $(2,2)$ gives $(1,2)$, already present.
• loops cause no issue.

So we can add up to two more pairs, but any addition must preserve transitivity.

We now test subsets of $S=\{(1,3),(2,1),(2,3),(3,1),(3,2)\}$ of size $0,1,$ or $2$.

4. Case-by-case counting

Case A: Add 0 extra pairs

Relation: $R=R_0.$ It is reflexive, transitive, and not symmetric since $(1,2)\in R$ but $(2,1)\notin R$.

So this gives 1 relation.

Case B: Add 1 extra pair

We test each of the 5 possibilities.

(i) Add $(1,3)$

Then $R=R_0\cup\{(1,3)\}$. No new transitivity requirement fails. Valid.

(ii) Add $(2,1)$

Now $(2,1)$ and $(1,2)$ imply by transitivity that $(2,2)$ is present, yes. But $(1,2)$ and $(2,1)$ imply $(1,1)$, yes. However $(2,1)$ and $(1,2)$ don't force anything new beyond loops. This relation is actually symmetric on pair $(1,2)$, but symmetry overall also requires reverse of every pair present. Since only non-diagonal pairs are $(1,2)$ and $(2,1)$, it becomes symmetric. So not allowed.

(iii) Add $(2,3)$

Then $(1,2)$ and $(2,3)$ imply $(1,3)$ must be in $R$, but it is not. So not transitive, not allowed.

(iv) Add $(3,1)$

Then $(3,1)$ and $(1,2)$ imply $(3,2)$ must be in $R$, but it is not. So not transitive, not allowed.

(v) Add $(3,2)$

No transitivity failure occurs:

• $(3,2)$ with $(2,2)$ gives $(3,2)$.
• $(3,3)$ with $(3,2)$ gives $(3,2)$.
• $(1,2)$ with anything starting at 2 only gives something if we had $(2,1)$ or $(2,3)$, which we do not. Valid.

Hence Case B gives 2 valid relations: $\{(1,3)\},\ \{(3,2)\}.$

Case C: Add 2 extra pairs

Choose 2 pairs from $S$. There are $\binom52=10$ possibilities. Check each.

1. $(1,3),(2,1)$

Since $(2,1)$ and $(1,3)$ imply $(2,3)$, which is absent. Not transitive.

2. $(1,3),(2,3)$

Since $(1,2)$ and $(2,3)$ imply $(1,3)$, which is present. No further issue. Not symmetric because $(2,1)$ absent. Valid.

3. $(1,3),(3,1)$

Then $(3,1)$ and $(1,3)$ imply $(3,3)$, present; $(1,3)$ and $(3,1)$ imply $(1,1)$, present. But $(3,1)$ and $(1,2)$ imply $(3,2)$, absent. So not transitive.

4. $(1,3),(3,2)$

Check:

• $(1,3)$ and $(3,2)$ imply $(1,2)$, present. No other issue. Valid.

5. $(2,1),(2,3)$

Then $(1,2)$ and $(2,1)$ imply $(1,1)$, present. Also $(1,2)$ and $(2,3)$ imply $(1,3)$, absent. So not transitive.

6. $(2,1),(3,1)$

Then $(2,1)$ and $(1,2)$ imply $(2,2)$, present. $(3,1)$ and $(1,2)$ imply $(3,2)$, absent. So not transitive.

7. $(2,1),(3,2)$

Then $(3,2)$ and $(2,1)$ imply $(3,1)$, absent. So not transitive.

8. $(2,3),(3,1)$

Then $(1,2)$ and $(2,3)$ imply $(1,3)$, absent. So not transitive.

9. $(2,3),(3,2)$

Then $(1,2)$ and $(2,3)$ imply $(1,3)$, absent. Also $(3,2)$ and $(2,3)$ imply $(3,3)$, present. But missing $(1,3)$ already fails. Not transitive.

10. $(3,1),(3,2)$

$(3,1)$ and $(1,2)$ imply $(3,2)$, present. No further issue. Valid.

Thus Case C gives 3 valid relations.

5. Total count

Total valid relations: $1+2+3=6.$

So the required number of relations is $\boxed{6}.$

6. Comparison with stored answer

Stored correct answer = $6$.

Our derived answer also is $6$, so they agree.