From the FBD of the first and third segments

$R_1 = H_1 = 20 ~ \text{kN}$

$R_4 = H_2 = \frac{1}{2}(6 \times 10) = 30 ~ \text{kN}$

From the FBD of the second segment

$\Sigma M_{R3} = 0$
$10R_2 + 4(30) = 14(20) + 14(20)(7)$

$R_2 = 212 ~ \text{kN}$

$\Sigma M_{R2} = 0$
$10R_3 + 4(20) = 14(30) + 14(20)(3)$

$R_3 = 118 ~ \text{kN}$