What are the equations giving the molar flow of different molecules linked by a reaction?
I can't tell you precisely where your reasoning starts to be wrong, but here is another way to think about the problem.

Notice that for a specific reaction as you gave here, the term $rV$ is always the same independently of the molecule you choose to determine it. It does vary for sure over time though (don't read I said it was a constant).

As you wrote:

$$\begin{cases} F_{A,in} + 0 - arV = F_{A,out} \\ F_{B,in} + 0 - brV = F_{B,out} \\ F_{C,in} + crV - 0 = F_{C,out} \\ \end{cases}$$

Thus, $$rV = \frac{F_{A,in} - F_{A,out}}{a} = \frac{F_{B,in} - F_{B,out}}{b} = \frac{F_{C,in} - F_{C,out}}{-c}$$

You can write what happens for $C$ in terms of $A$: $$\frac{F_{A,in} - F_{A,out}}{a} = \frac{F_{C,in} - F_{C,out}}{-c} \Leftrightarrow F_{C,out} = F_{C,in} + \frac{c}{a} \left(F_{A,in} - F_{A,out}\right)$$

Which boils down to, $$F_{C,out} = F_{C,in} + \frac{c}{a} X F_{A,in}$$

And if you write equations in function of a particular molecule, $A$, as you did here, make sure that the conversion you wrote $X$, is to be more precise, the conversion of $A$ at the exit of the reactor. Otherwise, it doesn't make sense.



EDIT:

Note as well that your equation for $B$ is wrong as well, using the same reasoning as before you should find: $$F_{B,out} = F_{B,in} - \frac{b}{a} X F_{A,in}$$ which make sense because the flow of $B$ must be less at the end of the reactor if this reaction really occurs.