$\dfrac{R_2 L^3}{48EI} = \dfrac{5wL^4}{384EI}$
$R_2 = \frac{5}{8}wL = \frac{5}{8}(21.5)(12)$
$R = 161.25 ~ \text{kN}$ ← Answer for Part (3)
$\Sigma F_v = 0$
$2R_1 + R_2 = 21.5(12)$
$2R_1 + 161.25 = 258$
$R_1 = 48.375 ~ \text{kN}$
$M_2 = 48.375(6) - 21.5(6)(3)$
$M_2 = -96.75 ~ \text{kN}\cdot\text{m}$ ← Answer for Part (1)
$V_x = 48.375 - 21.5x = 0$
$x = 2.25 ~ \text{m}$
$M_x = 48.375x - 21.5x\left( \dfrac{x}{2} \right)$
$M_x = 48.375(2.25) - 10.75(2.25^2)$
$M_x = 54.42 ~ \text{kN}\cdot\text{m}$ ← Answer for Part (2)
