In case you haven't worked it out yet, here's my best explanation.
Firstly, notice how we can have any number of $a$ characters at the start. So, to make life easy let's assume we have none. If we then let $w=b^pc^p$, we can see that this is part of $L$ (when $r=0$ and $q=p$).
We then take a look at the conditions for a regular language under the pumping lemma. If the language is regular, we can take any $w\in L$ and decompose it into $xyz$ where:
- $|y|\geq1$
- $|xy|<p$
- $xy^nz\in L\forall n\in\mathbb{N}_0$
If we go back to $w=b^pc^p$, we can see that because $|xy|\leq p$, $xy$ has to be made up of entirely the character $b$ otherwise $|xy|>p$. This means that $y$ must be made up of at least 1 $b$ (because $|y|\geq1$) and at most $p$ repetitions of $b$.
We then have a look at the last condition of the pumping lemma; $xy^nz\in L\forall n\in\mathbb{N}_0$. Since $y$ is made up of only repetitions of $b$, any repetition of $y$ will add more of the character $b$ into this new word. This means that $xy^nz\notin L$ when $n\neq 1$ and therefore the last condition of the pumping lemma is not met so $L$ is not a regular language.
Hope this was helpful in some way
