Is the following language is a context free grammar language?
Consider the word $w = 0^n1^n0^n \# 0^n1^n0^n \in L$ for a sufficiently large value of $n$.

By the pumping lemma for context-free languages, we know that if $L$ is context-free then there are$a,b,c,d,e = \Sigma^*$, with $|bcd| &lt; n$ and $|bd|&gt;1$ such that $w=abcde$ and, for every $i \ge 0$, $a b^i c d^i e\in L$.

Notice that neither $b$ nor $d$ can contain $\#$, as otherwise setting $i=0$ would show that $a c e \not\in L$.

If $c$ contains $\#$, then $b,d \in 0^*$ and $a b^i c d^i e$ is a word of the form $0^n 1^n 0^{x + i|b|}\#0^{y+i|d|} 1^n0^n$ with $x+|b|=n$ and $y+|d|=n$.  If $|b|&gt;0$ you can make the first part of the word (before $\#$) end with more than $n$ zeros by picking $i=2$, resulting in a contradiction. Otherwise $|d|&gt;0$ and you can make the second part of the word (after $\#$) start with less than $n$ zeros by setting $i=0$.

If $c$ does not contain $\#$ then $b$ and $c$ either both precede or both follow the unique occurrence of $\#$ in $w$. If they precede $\#$, pick $i=2$. If they follow $\#$, pick $i=0$.
In this way the first part of the word becomes longer than the second (and hence not in $L$).