By pumping lema, we have that for each sufficiently large $z \in L$, there exists $u, v$ and $w$ such as $z=uvw$, whith $|uv| \leq p$, $|v| \ge 1$ and $uv^iw \in L$, for each $i \ge 0.$ ($p$ is the threshold given by PL)
When I want to find a word that doesn't respect the PL, it must have some $a$-s as a prefix, because $k >0$. Given this impediment, I can not prove that any decomposition $u, v, w$ gives me a contradiction, because there it is always the possibility that $uv = a^j$ for some $j$ and this keeps $uv^i w$ in $L$.
What can I do? It works with proving that only the second part of the words is not regular and this will imply that the concatenation is also not regular?
HINT: If you know that the class of regular languages is closed under the reversal operator, you can use the pumping lemma to prove that the language
$$L'=\left\{c^\ell b^{3\ell}a^k:\ell\ge 0\text{ and }k>0\right\}$$
is not regular; since $L=\left\{w^R:w\in L'\right\}$, it follows that $L$ is not regular either.
Otherwise, you can use the Myhill-Nerode theorem by finding an infinite $D\subseteq\{a,b,c\}^*$ such that whenever $u,v\in D$ and $u\ne v$, the words $u$ and $v$ have a distinguishing extension. If you adopt this approach, I suggest trying to find one member of $D$ for each $\ell\ge 1$.