Correct Option: B

The context free grammar given over alphabets
∑ = {a, b, c}, with S and T are nonterminals.
Given, G₁ : S → aSb | T , T → cT | ∈
G₂ : S → bSa | T , T → cT | ∈
Lets L(G₁) is the language for Grammar (G₁) and L(G₂) is the language for Grammar (G₂).
where L(G₁) = { aⁿ c^m bⁿ | m , n ≥ 0 }
L(G₂) = { bⁿ c^m aⁿ | m , n ≥ 0 }
then L(G₁) ∩ L(G₂) = { aⁿ c^m bⁿ } ∩ { bⁿ c^m aⁿ }
= { c^m | m ≥ 0 } = C^*
Since the only common strings will be those strings with only ‘C’, so the intersection is not finite but regular.