If $S$ is the circumcentre,$G$ the centroid,and $O$ the orthocentre of a triangle $ABC$,then $\overrightarrow {SA} + \overrightarrow {SB} + \overrightarrow {SC} = $

Between the following two statements :
Statement $-I$ : Let $\vec{a}=\hat{i}+2\hat{j}-3\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$. Then the vector $\vec{r}$ satisfying $\vec{a} \times \vec{r}=\vec{a} \times \vec{b}$ and $\vec{a} \cdot \vec{r}=0$ is of magnitude $\sqrt{10}$.
Statement $-II$ : In a triangle $ABC$,$\cos 2A+\cos 2B+\cos 2C \geq -\frac{3}{2}$.

In the given figure,if a vector $x$ satisfies the equation $x - w = v$,then $x = ?$

If $a=2 \hat{i}+3 \hat{j}+\hat{k}$,$b=\hat{i}-3 \hat{j}-5 \hat{k}$ and $c=3 \hat{i}-4 \hat{k}$,then match the items of List-$I$ with those of List-$II$.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$,then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then,its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$,then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

The point $B$ divides the arc $AC$ of a quadrant of a circle in the ratio $1 : 2$. If $O$ is the centre and $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$,then the vector $\overrightarrow{OC}$ is

If $\bar{a}, \bar{b}, \bar{c}$ are three vectors such that $|\bar{a}|=\sqrt{31}, 4|\bar{b}|=|\bar{c}|=2$ and $2(\bar{a} \times \bar{b})=3(\bar{c} \times \bar{a})$ and if the angle between $\bar{b}$ and $\bar{c}$ is $\frac{2\pi}{3}$,then $\left|\frac{\bar{a} \times \bar{c}}{\bar{a} \cdot \bar{b}}\right|^2=$