Consider the context-free grammars over the alphabet $\{a,b,c\}$ given below. $S$ and $T$ are non-terminals. $G_1: S \to aSb | T, \ T \to cT | \epsilon$ $G_2: S \to bSa | T, \ T \to cT | \epsilon$ The language $L(G_1) \cap L(G_2)$ is

From the grammar $G_1$:

• $S \to aSb \mid T$ generates equal numbers of $a$'s at the beginning and $b$'s at the end surrounding $T$.

• $T \to cT \mid \epsilon$ generates any number of $c$'s ($c^*$).

• Thus, $L(G_1) = \{ a^n c^k b^n \mid n \ge 0, k \ge 0 \}$.

From the grammar $G_2$:

• $S \to bSa \mid T$ generates equal numbers of $b$'s at the beginning and $a$'s at the end surrounding $T$.

• $T \to cT \mid \epsilon$ generates $c^*$.

• Thus, $L(G_2) = \{ b^n c^k a^n \mid n \ge 0, k \ge 0 \}$.

To find the intersection $L(G_1) \cap L(G_2)$:

• For $n > 0$, any non-empty string generated with $n > 0$ in $L(G_1)$ starts with $a$ and ends with $b$, whereas in $L(G_2)$ it starts with $b$ and ends with $a$. They share no common strings for $n > 0$.

• For $n = 0$, both grammars generate $T$, which produces strings in $\{ c^k \mid k \ge 0 \} = c^*$.

Hence, $L(G_1) \cap L(G_2) = c^*$.

The language $c^*$ is infinite (not finite) and regular, which corresponds to option (B).