Since the constraint includes binaries, it does not define a convex set, and is therefore not linear.
For example, if $x=c_11_{A}$, $x$ can take values either $0$ or $c_1$. But $\frac{0+c_1}{2} \notin \{0,c_1 \}$, so the constraint does not define a convex set.
Note however that the convex hull of the constraint is linear, which is why such constraints are considered linear in the context of Mixed Linear Programming (MIP). So as Erwin points out in the comment section, a MIP is linear/convex if its relaxation is linear/convex.