Is $A=\{ w \in \{a,b,c\}^* \mid \#_a(w)+ 2\#_b(w) = 3\#_c(w)\}$ a CFG?
Yes, $A$ is a CFL.  Use the context free language (with the notation $|x|_0$ meaning what you have as #0's in $x$):

$B=\{x\in \{0,1\}^*:3\cdot|x|_0=|x|_1\}$ 

and the morphism $f:\{a,b,c\}^*\rightarrow\{0,1\}^*:$ $ f(a)=0, f(b)=00, f(c)=1$ 

so that 

$A=f^{-1}(B)$. 

Since the CFL's are closed under inverse morphism, $A$ is a CFL.

The proof that $B$ is a CFL is a bit tricky.  Let's do it for the simpler case $C=\{x:|x|_0=|x|_1\}$, which can fairly easily be generalized to work for $B$ (or use another morphism).  

Design a PDA which keeps track of $|x|_0-|x|_1$ at all points of the input, using the stack to keep track of the difference, having read $x$. The PDA accepts only when that difference is zero, that is, the stack is empty. The idea is to push a $0$ on seeing a $0$ on the input string, and pop a $0$ when seeing a $1$ in the input. The problem is that difference might go negative at times, and you can't pop an empty stack. In that case, however, the PDA goes into another mode (a different set of states) where it pushes a $1$ on seeing an input of $1$ and pops a $1$ on seeing an input of $0$.  So the second mode handles the negative case. It alternates between the two modes, accepting only if the stack if totally empty, that is, in-between the modes.

Or, you can do it with the grammar: $S\rightarrow 0S1|1S0|SS|\epsilon$.  That clearly generates only strings of C, but proving that it generates all of $C$ is a bit involved, but here's a sketch, by induction on string length.  

An isosymbolic string $x$ (that is $|x|_0 = |x|_1$) could be null, in which case the $\epsilon$ production applies giving the basis of the induction.  Otherwise $x$ either starts and ends in different symbols (one end $0$ and the other end $1$), or the same symbol (both $0$ or both $1$). In the not-equal case, the last production to apply is either the $0S1$ or the $1S0$ production, allowing us to strip off the first and last symbols, getting a shorter isosymbolic string, so the induction applies.  

In the equal case, as with the PDA, we keep track of $|y|_0-|y|_1$ as we move from left to right through prefixes $y$ of $x$. If the first and last symbols of $x$ are the same, it's fairly easy to see that that expression must be zero somewhere in the interior of the string. For example, if $x$ starts and ends with $0$, then after the first symbol the difference is $+1$ and before the last symbol it is $-1$, so it must have crossed zero in the interior. Split the string in two at that spot (reversing the $SS$ production) and the two halves, both shorter than $x$, are both isosymbolic.  So the induction works there too, and we are done.