1. Given relation

We have $A=\{1,2,3,4,5,6,7\}$ and $R=\{(x,y)\in A\times A: x+y=7\}.$

So the ordered pairs in $R$ are all pairs from $A$ whose sum is $7$.

Listing them: $R=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}.$

Note that $(7,0)$ is not included since $0\notin A$, and $(3.5,3.5)$ is not possible since $3.5\notin A$.

2. Check reflexivity

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

Here, $(x,x)\in R$ would require $x+x=7 \implies 2x=7 \implies x=\frac{7}{2},$ which is not an element of $A$.

So no pair of the form $(x,x)$ belongs to $R$. Hence, $R$ is not reflexive.

3. Check symmetry

A relation is symmetric if whenever $(x,y)\in R$, then $(y,x)\in R$.

If $(x,y)\in R$, then $x+y=7.$ But then $y+x=7,$ so $(y,x)\in R$ as well.

For example:

• $(1,6)\in R$ and $(6,1)\in R$
• $(2,5)\in R$ and $(5,2)\in R$
• $(3,4)\in R$ and $(4,3)\in R$

Therefore, $R$ is symmetric.

4. Check transitivity

A relation is transitive if whenever $(x,y)\in R$ and $(y,z)\in R$, then $(x,z)\in R$.

Take $(1,6)\in R \quad \text{and} \quad (6,1)\in R.$ If $R$ were transitive, then we must have $(1,1)\in R.$ But $1+1=2\neq 7,$ so $(1,1)\notin R$.

Hence, $R$ is not transitive.

5. Conclusion

Thus, the relation $R$ is:

• symmetric
• not reflexive
• not transitive

So the correct option is: $\boxed{\text{C: symmetric but neither reflexive nor transitive}}$

6. Comparison with stored answer

Stored correct answer: C

Our derived answer is also C, so they agree.