For the three-charge system to be in equilibrium, the net force on each charge must be zero.

1. Position of C: For charge C to be in equilibrium, it must be placed on the line connecting A and B. Since A and B have opposite signs, C must be placed outside the segment AB, on the side of the charge with the smaller magnitude, which is B ($-Q$). Let C be at a distance $x$ from B. Equilibrium of C: $F_{AC} = F_{BC}$

$$k \frac{|q_A q_C|}{(L+x)^2} = k \frac{|q_B q_C|}{x^2} \implies \frac{4Q}{(L+x)^2} = \frac{Q}{x^2}$$ $$4x^2 = (L+x)^2 \implies 2x = L+x \implies x=L$$

So, C is located at a distance L from B, away from A.

2. Charge of C: For charge B to be in equilibrium, the forces from A and C must cancel. Let A be at $x=0$, B at $x=L$, C at $x=2L$. The force on B from A ($F_{AB}$) is attractive, directed towards A ($-x$ direction). The force on B from C ($F_{CB}$) must be in the opposite ($+x$) direction. Since C is at $x=2L$, it must attract B. As B ($-Q$) is negative, C must be positive. Equilibrium of B: $|F_{AB}| = |F_{CB}|$

$$k \frac{|q_A q_B|}{L^2} = k \frac{|q_C q_B|}{x^2} \implies k \frac{(4Q)(Q)}{L^2} = k \frac{|q_C|(Q)}{L^2}$$ $$4Q = |q_C|$$

Since we determined $q_C$ must be positive, $q_C = +4Q$. We can verify that charge A is also in equilibrium with this configuration.