Equal No of a's ,b's and c's is not context-free, but it's complement is Context-free.
That is ,
$L$ = {$w$ |$no_a(w)\neq no_b(w)$,$w\in(a+b+c)^*$}
$\cup$
{$w$ |$no_b(w)\neq no_c(w)$,$w\in(a+b+c)^*$}
$\cup$
{$w$ |$no_c(w)\neq no_a(w)$,$w\in(a+b+c)^*$}
Note: I hope it can be implemented by NPDA