How to check if a language is not regular?
Yes your answer is correct.
Language L generates strings that begin with 2as followed by any number of bs then followed by any number of cs
Your regular expression represents L correctly
It is also worth reminding how the pumping lemma works , if a string in a language L cannot be pumped , then L is non-regular , however some languages can still fool the pumping lemma
Consider the language F = {a^i b^j c^k| i,j,k ≥ 0 and if i = 1 then j = k}. Which appears as a regular language in pumping lemma but is actually non-regular
This is why there are other methods to prove that a language is non-regular
For example to prove F is non-regular you should remember that regular languages are closed under complement (if F is regular then F' is regular too ) , then by the pumping lemma you can show that F' is non-regular and thus F is non-regular , sometimes closure under intersection is useful too
Finally you should try to get an intuition on the language , clearly L needs only finite memory to check the number of as , there can be any number of bs and cs and the FA does not need to remember the number of bs and cs , an intuition is a good first step .