From the FBD of larger pulley

$R_B = 2(10 \cos 60^\circ)$
$R_B = 10 ~ \text{kN}$

From the FBD of smaller pulley

$R_D = 2(10 \cos 30^\circ)$
$R_D = 10\sqrt{3} ~ \text{kN} = 17.32 ~ \text{kN}$

From the FBD of boom AD

$\Sigma M_A = 0$
$3T = 2(10)$

$T = 20/3 ~ \text{kN} = 6.67 ~ \text{kN}$

$\Sigma F_x = 0$

$A_H + R_B \cos 30^\circ = T \cos 30^\circ + R_D \cos 60^\circ$

$A_H + 10 cos 30^\circ = \frac{20}{3} \cos 30^\circ + 10\sqrt{3} \cos 60^\circ$

$A_H = \frac{10}{3}\sqrt{3} ~ \text{kN} = 5.77 ~ \text{kN}$

$\Sigma F_y = 0$
$A_V + T \sin 30^\circ = R_B \sin 30^\circ + R_D \sin 60^\circ$

$A_V + \frac{20}{3} \sin 30^\circ = 10 \sin 30^\circ + 10 \sin 60^\circ$

$A_V = 50/3 ~ \text{kN} = 16.67 ~ \text{kN}$

$R_A = \sqrt{\left( \frac{10}{3}\sqrt{3} \right)^2 + \left( \frac{50}{3} \right)^2}$

$R_A = \frac{20}{3}\sqrt{7} ~ \text{kN} = 17.64 ~ \text{kN}$ *answer*