Need some explanation of Pumping lemma for CFL I need some help with the understanding of Pumping Lemma for CFL 
L = {all words over $\{a,b,c\}$ s.t. $n_a=n_b+2n_c\}$ where $n_a$ stands for number of $a$,$n_b$ - number of $b$ and $n_c$ number of $c$.
show that $L$ is not $CFL$  
My thoughts:
Assume $L$ is $CFL$
take $w = b^{m}c^{2m}a^{3m}$.
$vy$ is either $b^{i} c^{j}$ or $c^{i} c^{j}$ or $c^{i} a^{j}$ or $a^{i} a^{j}$
take $w_1=uvvxyyz$
$w_1$ doesn't in $L$ since it doesn't satisfie the rule $n_a=n_b+2n_c $ 
Thus $L$ is not $CFL$
Can you show me where I'm wrong?  
 A: I hope it can help you.
Pumping lemma for CFL :
for infinite context-free language L 
there exists an integer $m$ such that:
for any string  $w\in L$ ,$|w| \ge m$
we can write $w=uvxy$


*

*$|vxy|\le m$ 

*$|vy| \ge 1$
and it must be:  $\,\,\,\,\,\,uv^ixy^iz\in L$  $\,\,\,\,\,\,\,\,\forall i \ge 0 $

The language $L$ is not context-free language 
Use the Pumping Lemma
for context-free languages .we proof by contradiction (assume L is CFL pick any string of L with length at least m) examine $\underline{all}$ the possible locations of string $vxy$ in string $w$ and show contradiction in each case  . 

A language is context-free if there is a CFG for it

A context-free grammar (CFG) consists of a set of productions that you
  use to replace a variable by a string of variables and terminals.
  The language of a grammar is the set of strings it generates. A
  language is context-free if there is a CFG for it

your language in question is context-free language.
for $L=\{w\in\{a,b,c\}^*: n_a=n_b+2n_c , n_a \ge 0,n_b \ge 0,n_c \ge 0\}$ we have this grammar :
$\,\,\,\,\,\,\,\,\,S\to aaSc|A$
 $\,\,\,\,\,\,\,\,\,A\to aAb|\epsilon$ 
so exist a context-free grammar that generating this language.
$\,\,\,\,\,\,\,\,$=> L is context-free language. 
