Given relation

Let $R=\{(x,y)\in W\times W \mid x \text{ and } y \text{ have at least one letter in common}\}$ where $W$ is the set of English words.

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

Reflexive check

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

Any word $x$ has all its own letters in common with itself, so certainly $x$ and $x$ have at least one letter in common.

Hence, $(x,x)\in R \quad \forall x\in W$

Therefore, $R$ is reflexive.

Symmetric check

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

If $x$ and $y$ have at least one letter in common, then clearly $y$ and $x$ also have that same letter in common.

So, $(x,y)\in R \implies (y,x)\in R$

Therefore, $R$ is symmetric.

Transitive check

A relation is transitive if $(x,y)\in R \text{ and } (y,z)\in R \implies (x,z)\in R.$

This need not be true here.

Take examples:

• $x = \text{"at"}$
• $y = \text{"tap"}$
• $z = \text{"pet"}$

Now:

• $x$ and $y$ have letters in common: $a$ or $t$
• $y$ and $z$ have letters in common: $p$ or $t$
• But let us choose a cleaner counterexample to avoid overlap confusion.

Better example:

• $x=\text{"ab"}$
• $y=\text{"bc"}$
• $z=\text{"cd"}$

Then:

• $x$ and $y$ have common letter $b$, so $(x,y)\in R$
• $y$ and $z$ have common letter $c$, so $(y,z)\in R$
• But $x$ and $z$ have no common letter, so $(x,z)\notin R$

Thus transitivity fails.

Therefore, $R$ is not transitive.

Conclusion

The relation $R$ is:

• reflexive
• symmetric
• not transitive

So the correct option is: $\boxed{\text{A}}$