Option (A): Let $\vec{a}=\alpha \hat{i}+\beta \hat{j}$ and $\vec{b}=\sqrt{3} \hat{i}+\hat{j}$.

Therefore, the magnitude of projection of $\vec{a}$ on $\vec{b}$ is

$$ \begin{aligned} &\vec{b}=\frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} \\\\ &=\frac{|\sqrt{3} \alpha+\beta|}{\sqrt{3+1}}=\sqrt{3} \\\\ &\Rightarrow \sqrt{3} \alpha+\beta= \pm 2 \sqrt{3} \\\\ &\Rightarrow \sqrt{3}(2+\sqrt{3} \beta)+\beta= \pm 2 \sqrt{2} \\\\ &\Rightarrow \beta=0 \text { or } \beta=-\sqrt{3} \Rightarrow \alpha=2 \text { or } \alpha=-1 \\\\ &\Rightarrow|\alpha|=2 \text { or } 1 \end{aligned} $$

Hence, $(\mathrm{A}) \rightarrow(\mathrm{P}),(\mathrm{Q})$.

Option (B): $f(x)=\left\{\begin{array}{r}-3 a x^2-2, x<1 \\ b x+a^2, x>1\end{array}\right.$

Since $f(x)$ is differentiable $\forall x \in \mathbb{R}$, we have $f\left(1^{-}\right)=f\left(1^{+}\right)$

$$ \begin{aligned} &\Rightarrow-3 a-2 =b+a^2 \\\\ &\Rightarrow a^2+3 a+2 =-b \\\\ &\Rightarrow(a+2)(a+1) =-b \quad\quad...(1) \end{aligned} $$

Also, $f^{\prime}(x)=\left\{\begin{array}{cc}-6 a x ; & x<1 \\ b ; & x>1\end{array}\right.$

$$ =f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right) $$

$$ \Rightarrow \quad-6 a=b \quad\quad...(2) $$

Therefore, from Eqs. (1) and (2), we get (2) $a^2+3 a+2$ $=6 a$

$$ \Rightarrow a=1 \text { or } a=2 $$

Hence, (B) $\rightarrow(\mathrm{P})$, (Q).

Also,

$$ \begin{aligned} &f^{\prime}(x) =\left\{\begin{array}{cc} -6 a x, & x<1 \\ b, & x \geq 1 \end{array}\right. \\\\ &\Rightarrow f^{\prime}\left(1^{-}\right) =f^{\prime}\left(1^{+}\right) \\\\ &\Rightarrow-6 a =b \\\\ &f^{\prime}\left(1^{-}\right) =f^{\prime}\left(1^{+}\right), a^2+3 a+2=6 a \\\\ &\Rightarrow a =1 \text { or } a=2 \end{aligned} $$

Hence, $(\mathrm{B}) \rightarrow(\mathrm{P}),(\mathrm{Q})$

$$ \begin{aligned} & \mathbb{\text {Option (C): }}\left(3-3 \omega+2 \omega^2\right)^{4 x+3}+\left(2+3 \omega-3 \omega^2\right)^{4 x+3}+ \\\\ & \left(-3+2 \omega+3 \omega^2\right)^{4 x+3}=0 \\\\ & \Rightarrow\left[1-3 \omega+2\left(1+\omega^2\right)\right]^{4 x+3}+\left[2(1+\omega)+\omega-3 \omega^2\right]+ \\\\ & {\left[-3+\omega^2+2\left(\omega+\omega^2\right)\right]^{4 x+3}=0} \end{aligned} $$

$$ \begin{aligned} & \Rightarrow {[1-3 \omega-2 \omega]^{4 x+3}+\left[-5 \omega^2+\omega\right]^{4 x+3} } \\\\ & \quad \quad\left(\omega^2\right)^{4 x+3}(1-5 \omega)^{4 x+3}=0 \\\\ & \Rightarrow(1-5 \omega)^{4 x+3}\left(1+\omega^n+\omega^{2 x}\right)=0 \\\\ & \Rightarrow \omega(1-5 \omega)^{4 x+3} \neq 0 \\\\ & \Rightarrow 1+\omega^x+\omega^{2 x}=0 \\\\ & \Rightarrow x=3 k+1 \text { or } x=3 k+2 ; k \in \mathbb{Z} \\\\ & \Rightarrow x \in\{1,2,4,5\} \end{aligned} $$

Hence, $(\mathrm{C}) \rightarrow(\mathrm{P}),(\mathrm{Q}),(\mathrm{S}),(\mathrm{T})$.

Option (D): HM of $a$ and $b=\frac{2 a b}{a+b}=4$, where $a, b>0$. Now, $a, 5, q, b$ are in AP, where $q>0$.

$$ \begin{gathered} \Rightarrow a+b=5+q \\\\ \Rightarrow \frac{a b}{2}=5+q ~~~~~~~~(1) \end{gathered} $$

Also

$$ \begin{aligned} &a+q= 10 \text { and } q=\frac{5+b}{2} ~~~~~~~~(2)\\\\ & \Rightarrow b=2 q-5(3) \end{aligned} $$

Therefore, from Eqs. (1), (2) and (3), $a=\frac{5}{2}$ or $a=6$.

$$ \Rightarrow q=\frac{15}{2} \text { or } 4 \Rightarrow|q-a|=5 \text { or } 2 $$

Hence, (D) $\rightarrow$ (Q), (T).