$\Sigma F_v = 0$
$3R = 300(2x + 6)$
$R = 200(x + 3) ~ \text{N}$
$\begin{align}
M_B & = \Sigma M_{\text{Left of } B} \\
& = 3R - 300(x + 3)\left( \dfrac{x + 3}{2} \right) \\
& = 600(x + 3) - 150(x + 3)^2 ~ \text{N}\cdot\text{m}
\end{align}$
$\begin{align}
M_A = M_C & = -300x \left( \dfrac{x}{2} \right) \\
& = -150x^2 ~ \text{N}\cdot\text{m}
\end{align}$
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)
