The balance equations would be the following

$$\Sigma F_x = A_x-C_x+\frac{G}{\sqrt{2}}=0$$
$$\Sigma F_y = A_y+C_y+\frac{G}{\sqrt{2}}=0$$
$$\Sigma M_P = A_y a-C_y a=0$$
$$\Sigma F_x = B_x+C_x-\frac{G}{\sqrt{2}}=0$$
$$\Sigma F_y = B_y-C_y-\left(\frac{G}{\sqrt{2}}+G\right)=0$$
$$\Sigma M_B = C_x (3 a)-\frac{G}{\sqrt{2}}(7 a)+
   \left(\frac{G}{\sqrt{2}}+G\right)(3 a)=0$$

And the solution turns out as
$$ A_x=\frac{1}{6} \left(\sqrt{2} G-6 G\right) \ \ A_y=-\frac{G}{2
   \sqrt{2}}$$
$$B_x=\frac{1}{6} \left(6 G-\sqrt{2} G\right)\ \ B_y=\frac{1}{4} \left(\sqrt{2}
   G+4 G\right)$$
$$C_x=\frac{1}{3} \left(2 \sqrt{2} G-3 G\right) \ \ C_y=-\frac{G}{2
   \sqrt{2}}$$

The free-body diagrams

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/h5OB8.png