Let $A = \{1, 2, 3\}$. We are looking for relations on $A$ that contain $(1, 2)$ and $(2, 3)$, are reflexive and transitive, but not symmetric.

Since the relation is reflexive, it must contain $(1, 1)$, $(2, 2)$, and $(3, 3)$.

Since the relation contains $(1, 2)$ and $(2, 3)$ and is transitive, it must also contain $(1, 3)$.

So, the relation must contain the following pairs: $(1, 1)$, $(2, 2)$, $(3, 3)$, $(1, 2)$, $(2, 3)$, $(1, 3)$.

Now, we need to consider additional pairs that can be added to the relation without making it symmetric.

If we add $(2, 1)$, the relation becomes symmetric because we already have $(1, 2)$. If we add $(3, 2)$, the relation becomes symmetric because we already have $(2, 3)$. If we add $(3, 1)$, the relation becomes symmetric because we already have $(1, 3)$.

So, we can consider adding $(2, 1)$, $(3, 2)$, and $(3, 1)$ individually or in combinations.

Case 1: Add $(2, 1)$. The relation is $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1)\}$. This is not transitive because $(2, 3)$ and $(3, 1)$ are in the relation, but $(2, 1)$ is. However, $(2,1)$ and $(1,3)$ are in the relation, but $(2,3)$ is. Also $(2,1)$ and $(1,2)$ are in the relation. This relation is not symmetric because $(2,3)$ is in the relation but $(3,2)$ is not. It is not transitive because $(2,1)$ and $(1,3)$ are in the relation but $(2,3)$ is. So this case is not valid.

Case 2: Add $(3, 2)$. The relation is $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 2)\}$. This is not symmetric because $(1, 2)$ is in the relation but $(2, 1)$ is not. It is transitive because $(3, 2)$ and $(2, 3)$ are in the relation, so $(3, 3)$ is in the relation. Also, $(1, 2)$ and $(2, 3)$ are in the relation, so $(1, 3)$ is in the relation. Also, $(3, 2)$ and $(2, 2)$ are in the relation, so $(3, 2)$ is in the relation. Also, $(1, 3)$ and $(3, 2)$ are not in the relation. So this case is valid.

Case 3: Add $(3, 1)$. The relation is $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 1)\}$. This is not symmetric because $(1, 2)$ is in the relation but $(2, 1)$ is not. It is not transitive because $(2, 3)$ and $(3, 1)$ are in the relation, but $(2, 1)$ is not. So this case is not valid.

Case 4: Add $(2, 1)$ and $(3, 2)$. The relation is $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1), (3, 2)\}$. This is not transitive because $(3, 2)$ and $(2, 1)$ are in the relation, but $(3, 1)$ is not. So this case is not valid.

Case 5: Add $(2, 1)$ and $(3, 1)$. The relation is $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1), (3, 1)\}$. This is not transitive because $(2, 3)$ and $(3, 1)$ are in the relation, but $(2, 1)$ is not. So this case is not valid.

Case 6: Add $(3, 2)$ and $(3, 1)$. The relation is $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 2), (3, 1)\}$. This is transitive. It is not symmetric. So this case is valid.

Case 7: Add $(2, 1)$, $(3, 2)$, and $(3, 1)$. The relation is $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1), (3, 2), (3, 1)\}$. This is not transitive because $(3, 2)$ and $(2, 1)$ are in the relation, but $(3, 1)$ is. So this case is not valid.

The valid cases are Case 2 and Case 6. Therefore, there is only 1 such relation.