$\Sigma F_v = 0$

$3R = 300(2x + 6)$

$R = 200(x + 3) ~ \text{N}$

$\begin{align}

M_B & = \Sigma M_{\text{Left of } B} \\

& = 3R - 300(x + 3)\left( \dfrac{x + 3}{2} \right) \\

& = 600(x + 3) - 150(x + 3)^2 ~ \text{N}\cdot\text{m}

\end{align}$

$\begin{align}

M_A = M_C & = -300x \left( \dfrac{x}{2} \right) \\

& = -150x^2 ~ \text{N}\cdot\text{m}

\end{align}$

Apply Three-Moment Equation to Span *A-B-C*

$M_A L_{AB} + 2M_B(L_{AB} + L_{BC}) + M_C L_{BC} + \left( \dfrac{6A\bar{a}}{L} \right)_{AB} + \left( \dfrac{6A\bar{b}}{L} \right)_{CB} = 0$

$-150x^2 (3) + 2M_B(3 + 3) - 150x^2 (3) + \dfrac{300(3^3)}{4} + \dfrac{300(3^3)}{4} = 0$

$12M_B = 900x^2 + 4050$

$M_B = 75x^2 - 337.5 ~ \text{N}\cdot\text{m}$

Hence,

$M_B = M_B$

$600(x + 3) - 150(x + 3)^2 = 75x^2 - 337.5$

$225x^2 + 300x - 787.5 = 0$

$x = 1.319 ~ \text{m}$ ← answer for Part (1)

$M_B = 75(1.319^2) - 337.5$

$M_B = 207 ~ \text{N}\cdot\text{m}$ ← answer for Part (2)

$R = 100 \, [ \, 2(1.319) + 6 \, ]$

$R = 863.8 ~ \text{N}$ ← answer for Part (3)