Answer
$2q,q,-3q$
Work Step by Step
We know that the net electric flux through the closed surface is given by
$$\Phi=\dfrac{Q_{in}}{\epsilon_0}$$
And from the given figures, we can see that
$$Q_{in,A}=q_1+q_3=-q$$
$$q_1+q_3=-q\tag 1$$
$$Q_{in,B}=q_1+q_2=3q$$
$$q_1+q_2=3q\tag 2$$
$$Q_{in,C}=q_2+q_3=-2q$$
$$q_2+q_3=-2q\tag 3$$\
From (2), $q_2=3q-q_1$ Plug that into (3),
$$(3q-q_1)+q_3=-2q$$
Hence,
$$ -q_1+q_3=-5q$$
Add this to (1),
$$q_1+q_3+(-q_1+q_3)=-q+(-5q)=-6q$$
Hence,
$$\boxed{q_3=-3q}$$
Plug into (2) to find $q_2$, and into (1) to find $q_1$
$$q_2-3q=-2q$$
Hence,
$$\boxed{q_2= q}$$
$$q_1-3q=-q$$
Hence,
$$\boxed{q_1= 2q}$$