$\overrightarrow{\mathrm{r}}\left(\mathrm{t}\right)=(1+2{\cos }^2\mathrm{\omega t})\widehat{\mathrm{i}}+\left(3{\sin }^2\mathrm{\omega t}\right)\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}$

$\overrightarrow{\mathrm{r}}\left(0\right)=3\widehat{\mathrm{i}}+3\widehat{\mathrm{k}}$

$\overrightarrow{\mathrm{s}}=\overrightarrow{\mathrm{r}}\left(\mathrm{t}\right)-\overrightarrow{\mathrm{r}}\left(0\right)=(2{\cos }^2\mathrm{\omega t}-2)\widehat{\mathrm{i}}+\left(3{\sin }^2\mathrm{\omega t}\right)\widehat{\mathrm{j}}$

$\overrightarrow{\mathrm{v}}=\frac{\mathrm{d}\overrightarrow{\mathrm{s}}}{\mathrm{dt}}=-4\mathrm{\omega sin\omega t}\widehat{\mathrm{i}}+3\mathrm{\omega sin}2\mathrm{\omega t}\widehat{\mathrm{j}}$

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d}\overrightarrow{\mathrm{V}}}{\mathrm{dt}}=-4{\mathrm{\omega }}^2\mathrm{cos\omega t}\widehat{\mathrm{i}}+6{\mathrm{\omega }}^2\cos 2\mathrm{\omega t}+\widehat{\mathrm{j}}$

$\overrightarrow{\mathrm{s}}=-4{\sin }^2\mathrm{\omega t}\widehat{\mathrm{i}}+3{\sin }^2\mathrm{\omega t}\widehat{\mathrm{j}}$

$\overrightarrow{\mathrm{v}}=(-4\mathrm{\omega sin}2\mathrm{\omega t})\widehat{\mathrm{i}}+\left(3\mathrm{\omega sin}2\mathrm{\omega t}\right)\widehat{\mathrm{j}}$

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d}\overrightarrow{\mathrm{V}}}{\mathrm{dt}}=\left(-8{\mathrm{\omega }}^2\cos 2\mathrm{\omega t}\right)\widehat{\mathrm{i}}+\left(6{\mathrm{\omega }}^2\cos 2\mathrm{\omega t}\right)\widehat{\mathrm{j}}$

$\widehat{\mathrm{a}}=\widehat{\mathrm{v}}$

$\left|\overrightarrow{\mathrm{a}}\right|=10{\mathrm{\omega }}^2\cos 2\mathrm{\omega t}$

$\left|\overrightarrow{\mathrm{v}}\right|=5\mathrm{\omega win}2\mathrm{\omega t}$

$\left|\overrightarrow{\mathrm{s}}\right|=5{\sin }^2\mathrm{\omega t}=\frac{5}{2}(1-\cos 2\mathrm{\omega t})$

$\left|\overrightarrow{\mathrm{a}}\right|=10{\mathrm{\omega }}^2\left[1-\frac{2}{3}\left|\overrightarrow{\mathrm{s}}\right|\right]$

$\mathrm{a}=10{\mathrm{\omega }}^2-4{\mathrm{\omega }}^2\mathrm{s}$

$\mathrm{a}={\mathrm{c}}_1-{\mathrm{c}}_2{\mathrm{\omega }}^2\mathrm{x}$

This is SHM with the mean position shifted from origin.

For mean position,

${\mathrm{V}}_{max.}=5\mathrm{\omega }\text{ at }\mathrm{t}=\frac{\mathrm{\pi }}{4\mathrm{\omega }}$

$\overrightarrow{\mathrm{s}}\left(\frac{\mathrm{\pi }}{4\mathrm{\omega }}\right)=1\widehat{\mathrm{i}}+\frac{3}{2}\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}\Rightarrow \left(1,\frac{3}{2},3\right)$

$10{\mathrm{\omega }}^2={\mathrm{A\omega }}_1^2$       .... (1)                      $5\mathrm{\omega }={\mathrm{A\omega }}_1$

Dividing  (2) / (1) :                                $25{\mathrm{\omega }}^2={\mathrm{A}}^2{\mathrm{\omega }}_1^2$         .... (2)

$\mathrm{A}=\frac{25}{10}=\frac{5}{2}\mathrm{m}$

$\widehat{\mathrm{v}}=\frac{-4{\displaystyle \widehat{\mathrm{i}}}}{3}+\frac{3{\displaystyle \widehat{\mathrm{j}}}}{5}\text{ or }\frac{4{\displaystyle \widehat{\mathrm{i}}}}{5}-\frac{3{\displaystyle \widehat{\mathrm{j}}}}{5}$

→ if a frame is moving with 3 m/s along z-axis, then $\overrightarrow{\mathrm{V}}$ will not be in x-y plane whereas $\overrightarrow{\mathrm{a}}$ is in x-y plane. So, not SHM.