Answer
$$
\begin{aligned}
& \mathbf{F}_R=\{141 \mathbf{i}+100 \mathbf{j}+159 \mathbf{k}\} \mathbf{N} \\
& \mathbf{M}_{R_O}=\{122 \mathbf{i}-183 \mathbf{k}\} \mathbf{N} \cdot \mathrm{m}
\end{aligned}
$$
Work Step by Step
Force And Moment Vectors:
$$
\begin{aligned}
\mathbf{F}_1 & =\{300 \mathbf{k}\} \mathbf{N} \quad \mathbf{F}_3=\{100 \mathbf{j}\} \mathbf{N} \\
\mathbf{F}_2 & =200\left\{\cos 45^{\circ} \mathbf{i}-\sin 45^{\circ} \mathbf{k}\right\} \mathbf{N} \\
& =\{141.42 \mathbf{i}-141.42 \mathbf{k}\} \mathbf{N} \\
\mathbf{M}_1 & =\{100 \mathbf{k}\} \mathbf{N} \cdot \mathrm{m} \\
\mathbf{M}_2 & =180\left\{\cos 45^{\circ} \mathbf{i}-\sin 45^{\circ} \mathbf{k}\right\} \mathbf{N} \cdot \mathrm{m} \\
& =\{127.28 \mathbf{i}-127.28 \mathbf{k}\} \mathrm{N} \cdot \mathrm{m}
\end{aligned}
$$
Equivalent Force and Couple Moment At Point O:
$$
\begin{aligned}
\mathbf{F}_R=\Sigma \mathbf{F} ; \quad \mathbf{F}_R & =\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3 \\
& =141.42 \mathbf{i}+100.0 \mathbf{j}+(300-141.42) \mathbf{k} \\
& =\{141 \mathbf{i}+100 \mathbf{j}+159 \mathbf{k}\} \mathbf{N}
\end{aligned}
$$
The position vectors are $\mathbf{r}_1=\{0.5 \mathbf{j}\} \mathrm{m}$ and $\mathbf{r}_2=\{1.1 \mathbf{j}\} \mathrm{m}$.
$$
\begin{aligned}
\mathbf{M}_{R_o}=\Sigma \mathbf{M}_o ; \quad \mathbf{M}_{R_o}= & \mathbf{r}_1 \times \mathbf{F}_1+\mathbf{r}_2 \times \mathbf{F}_2+\mathbf{M}_1+\mathbf{M}_2 \\
= & \left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 0.5 & 0 \\
0 & 0 & 300
\end{array}\right| \\
& +\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 1.1 & 0 \\
141.42 & 0 & -141.42
\end{array}\right| \\
& +100 \mathbf{k}+127.28 \mathbf{i}-127.28 \mathbf{k} \\
= & \{122 \mathbf{i}-183 \mathbf{k}\} \mathrm{N} \cdot \mathrm{m}
\end{aligned}
$$
