$\Sigma M_G = 0$
$12R_C + 6(60) = 6(100) + 9(80) + 18(40)$
$R_C = 140 ~ \text{kN}$
At section through M-M
$\Sigma F_V = 0$
$\frac{2}{\sqrt{5}}F_{BC} = 40$
$F_{BC} = 44.721 ~ \text{kN}$ compression answer
At section through N-N
$\Sigma M_D = 0$
$6F_{CE} + 9(40) = 3(140)$
$F_{CE} = 10 ~ \text{kN}$ tension answer
$\Sigma M_E = 0$
$6F_{DF} + 3(80) + 12(40) = 6(140) + 6(60)$
$F_{DF} = 80 ~ \text{kN}$ compression answer