Determine if the relation 'shares at least one letter' on a set of words is reflexive, symmetric, and transitive.

A relation $R$ on a set $W$ is reflexive if for every $x \in W$, $(x, x) \in R$.
For the given relation, $(x, x) \in R$ means that the word $x$ has at least one letter in common with itself. This is always true, as any word shares all its letters with itself.
Therefore, $R$ is reflexive.

A relation $R$ on a set $W$ is symmetric if for every $x, y \in W$, if $(x, y) \in R$, then $(y, x) \in R$.
If $(x, y) \in R$, it means that words $x$ and $y$ have at least one letter in common. If $x$ and $y$ have a common letter, then $y$ and $x$ also have that same letter in common.
Therefore, $R$ is symmetric.

A relation $R$ on a set $W$ is transitive if for every $x, y, z \in W$, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.
Let's test this with an example:
Let $x = \text{CAT}$
Let $y = \text{APPLE}$
Let $z = \text{DOG}$

• $(x, y) \in R$ because CAT and APPLE share the letter 'A'.
• $(y, z) \in R$ because APPLE and DOG share the letter 'P'.
• Now, we check if $(x, z) \in R$. Do CAT and DOG have any letters in common? No. Since $(x, z) \notin R$, the relation is not transitive.

The relation $R$ is reflexive, symmetric, and not transitive.

Based on the analysis, option (C) states "reflexive, symmetric, & not transitive", which matches our findings.

The final answer is $\text{C}$