You cannot find $c$. Eigenvectors are not unique. If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $cv$ is an eigenvector of $A$ for any $c\
eq 0$ with the same eigenvalue since
$$ A(cv) = c\cdot Av = c\cdot \lambda v \Longrightarrow \lambda (cv).$$
At best you can determine $v$ up to a constant multiple. You _could_ get a unique eigenvector if you required it to be a unit vector, but this only works if the eigenvalues have geometric multiplicity one. Otherwise there is inherent non-uniqueness since the eigenvalue (by definition of having geometric multiplicity greater than one) would correspond to multiple linearly independent eigenvectors.
I struggled with this myself for a while in undergrad. It's probably not that uncommon a mental disconnect, especially when first learning linear algebra.