Let $\Sigma = \{a, b, c\}$. For $x \in \Sigma^*$, and $\alpha \in \Sigma$, let $\#_\alpha(x)$ denote the number of occurrences of $\alpha$ in $x$. Which one or more of the following option(s) define(s) regular language(s)?

Let us analyze each of the given languages:

• Option (A): The language is $\{a^m b^n \mid m, n \ge 0\}$, which corresponds directly to the regular expression $a^*b^*$. Since it can be generated by a regular expression, it is regular.

• Option (B): The language is $\{a, b\}^* \cap \{a^m b^n c^{m-n} \mid m \ge n \ge 0\}$. Any string in $\{a^m b^n c^{m-n} \mid m \ge n \ge 0\}$ that contains no occurrences of the symbol $c$ must satisfy $m - n = 0 \implies m = n$. Thus, the intersection evaluates to $\{a^n b^n \mid n \ge 0\}$, which is well-known to be non-regular (requires unbounded memory to match the counts of $a$ and $b$).

• Option (C): The language is $\{w \mid w \in \{a, b\}^*, \#_a(w) \equiv 2 \pmod 7 \text{ and } \#_b(w) \equiv 3 \pmod 9\}$. This is the intersection of two regular languages (one keeping track of the count of $a$'s modulo $7$, requiring $7$ states, and the other keeping track of the count of $b$'s modulo $9$, requiring $9$ states). A product DFA with $7 \times 9 = 63$ states can recognize it, so it is regular.

• Option (D): The language requires strings over $\{a, b\}$ to satisfy both $\#_a(w) \equiv 2 \pmod 7$ and $\#_a(w) = \#_b(w)$. If we intersect this language with the regular language $a^* b^*$, we get $\{a^k b^k \mid k \ge 0 \text{ and } k \equiv 2 \pmod 7\}$. Using the Pumping Lemma, this language is non-regular; hence the original language is not regular.

Therefore, the correct options are A and C.