1. Set up the problem

Let $A=\{1,2,3\}.$ A relation on $A$ is any subset of $A\times A$.

We need relations $R$ such that:

• $(1,2)\in R$
• $(2,3)\in R$
• $R$ is reflexive
• $R$ is transitive
• $R$ is not symmetric

2. Use reflexivity

Since $R$ is reflexive, it must contain: $(1,1), (2,2), (3,3).$

Also given: $(1,2), (2,3) \in R.$

So currently, $R$ must contain at least $\{(1,1),(2,2),(3,3),(1,2),(2,3)\}.$

3. Use transitivity

Because $(1,2)\in R$ and $(2,3)\in R$, transitivity forces $(1,3)\in R.$

So now the relation must contain $\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}.$

4. Find the only pairs still optional

All possible ordered pairs in $A\times A$ are:

Out of these, the first six are already forced. Hence the only pairs whose inclusion/exclusion is undecided are: $(2,1), (3,1), (3,2).$

We now test all subsets of these three pairs, keeping transitivity valid.

5. Case analysis

Let the forced set be $R_0=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}.$

We may add any subset of $\{(2,1),(3,1),(3,2)\}$, but transitivity may force more additions.

Case 1: Add none

$R=R_0.$ Check transitivity: it already works. Also it is not symmetric because $(1,2)\in R$ but $(2,1)\notin R$.

So this gives 1 valid relation.

Case 2: Add only $(2,1)$

Then $R=R_0\cup\{(2,1)\}.$ Now from $(2,1)$ and $(1,3)$, transitivity requires $(2,3)$, already present. From $(1,2)$ and $(2,1)$, transitivity requires $(1,1)$, already present. From $(2,1)$ and $(1,2)$, transitivity requires $(2,2)$, already present. So transitivity still holds.

This relation is still not symmetric, since $(2,3)\in R$ but $(3,2)\notin R$.

So this gives 1 valid relation.

Case 3: Add only $(3,2)$

Then $R=R_0\cup\{(3,2)\}.$ From $(1,3)$ and $(3,2)$, transitivity requires $(1,2)$, already present. From $(2,3)$ and $(3,2)$, transitivity requires $(2,2)$, already present. From $(3,2)$ and $(2,3)$, transitivity requires $(3,3)$, already present. So transitivity holds.

This relation is not symmetric, since $(1,2)\in R$ but $(2,1)\notin R$.

So this gives 1 valid relation.

Case 4: Add only $(3,1)$

Then $R=R_0\cup\{(3,1)\}.$ Now $(2,3)$ and $(3,1)$ imply $(2,1)$ must be in $R$, but it is not. So transitivity fails.

Hence this case is invalid.

Case 5: Add $(2,1)$ and $(3,2)$

Then $R=R_0\cup\{(2,1),(3,2)\}.$ Now $(3,2)$ and $(2,1)$ imply $(3,1)$ must be in $R$. But it is not included, so transitivity fails.

Hence this case is invalid unless $(3,1)$ is also added.

Case 6: Add $(2,1)$ and $(3,1)$

Then $R=R_0\cup\{(2,1),(3,1)\}.$ Check transitivity:

• $(3,1)$ and $(1,2)$ imply $(3,2)$ must be in $R$. But $(3,2)$ is absent. So transitivity fails.

Hence this case is invalid.

Case 7: Add $(3,1)$ and $(3,2)$

Then $R=R_0\cup\{(3,1),(3,2)\}.$ Check transitivity:

• $(2,3)$ and $(3,1)$ imply $(2,1)$ must be in $R$. But $(2,1)$ is absent. So transitivity fails.

Hence this case is invalid.

Case 8: Add all three: $(2,1),(3,1),(3,2)$

Then $R=A\times A.$ This is reflexive, transitive, and symmetric. But we need not symmetric. So this case is invalid.

6. Count valid relations

The valid cases are only:

• $R_0$
• $R_0\cup\{(2,1)\}$
• $R_0\cup\{(3,2)\}$

Therefore, the number of such relations is $\boxed{3}.$

7. Compare with stored answer

Stored correct answer: $3$

Our derived answer is also $3$, so they agree.