Answer
Ans:
$$
\begin{aligned}
& \mathbf{v}_C=\{-4.8 \mathbf{i}-3.6 \mathbf{j}-1.2 \mathbf{k}\} \mathrm{m} / \mathrm{s} \\
& \mathbf{a}_C=\{38.4 \mathbf{i}-64.8 \mathbf{j}+40.8 \mathbf{k}\} \mathrm{m} / \mathrm{s}^2
\end{aligned}
$$
Work Step by Step
We need to express $w$ of the plate in vectorial form The unit vector that defined the direction of $\omega$ is?
$$
\mathbf{u}_{O A}=\frac{-0.3 \mathbf{i}+0.2 \mathbf{j}+0.6 \mathbf{k}}{\sqrt{(-0.3)^2+0.2^2+0.6^2}}=-\frac{3}{7} \mathbf{i}+\frac{2}{7} \mathbf{j}+\frac{6}{7} \mathbf{k}
$$
Then:
$$
\begin{aligned}
& \omega=\omega \mathbf{u}_{O A}=14\left(-\frac{3}{7} \mathbf{i}+\frac{2}{7} \mathbf{j}+\frac{6}{7} \mathbf{k}\right)=[-6 \mathbf{i}+4 \mathbf{j}+12 \mathbf{k}] \mathrm{rad} / \mathrm{s} \\
& \omega=\operatorname{cte}, \alpha=0 \\
& r_c=[-0.3 \mathbf{i}+0.4 \mathbf{j}] \mathrm{m}
\end{aligned}
$$
$$
\begin{aligned}
\mathbf{v}_C & =\omega \times \mathbf{r}_C \\
& =(-6 \mathbf{i}+4 \mathbf{j}+12 \mathbf{k}) \times(-0.3 \mathbf{i}+0.4 \mathbf{j}) \\
& =[-4.8 \mathbf{i}-3.6 \mathbf{j}-1.2 \mathbf{k}] \mathrm{m} / \mathrm{s}
\end{aligned}
$$
So:
$$
\begin{aligned}
\mathbf{a}_C & =\alpha \times \mathbf{r}_C+\omega \times\left(\boldsymbol{\omega} \times \mathbf{r}_C\right) \\
& =0+(-6 \mathbf{i}+4 \mathbf{j}+12 \mathbf{k}) \times[(-6 \mathbf{i}+4 \mathbf{j}+12 \mathbf{k}) \times(-0.3 \mathbf{i}+0.4 \mathbf{j})] \\
& =[38.4 \mathbf{i}-64.8 \mathbf{j}+40.8 \mathbf{k}] \mathrm{m} / \mathrm{s}^2
\end{aligned}
$$
