# Prove {0^n OR 1^2n OR 2^3n | n >= 0} is not context free

How to prove using pumping lemma {0^n OR 1^2n OR 2^3n | n >= 0} is not context free

This isnt the same language as {0^n1^2n2^3n | n >= 0} as this language the numbers need to be in order.

It is not specified but I'll assume the alphabet to be $\Sigma = \{0,1,2\}$ and also that your strange notation means the language
But it is also Regular! By definition a language is said to be regular if it is accepted by some $DFA$ or, equivalently, if it is expressible by a regular expression.
In our case, $L = 0^*+(11)^*+(222)^*$.