The vectors $\overrightarrow{AB} = 3\hat{i} + 5\hat{j} + 4\hat{k}$ and $\overrightarrow{AC} = 5\hat{i} - 5\hat{j} + 2\hat{k}$ are the sides of a triangle $ABC$. The length of the median through $A$ is ............. $unit$.

The vector $\frac{1}{3}(2i - 2j + k)$ is

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

If $\vec{a}+l \vec{b}+l^2 \vec{c}=0$ and $\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}=3(\vec{b} \times \vec{c})$, then the minimum value of such $l$ is

Let $A$ and $B$ be two points. The position vector of $A$ is $6b - 2a$. Point $P$ divides the line segment $AB$ in the ratio $1 : 2$. If $a - b$ is the position vector of $P$,what is the position vector of $B$?

Let $\vec{w}=\hat{i}+\hat{j}-2 \hat{k}$,and $\vec{u}$ and $\vec{v}$ be two vectors such that $\vec{u} \times \vec{v}=\vec{w}$ and $\vec{v} \times \vec{w}=\vec{u}$. Let $\alpha, \beta, \gamma$ and $t$ be real numbers such that $\vec{u}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$,$-t \alpha+\beta+\gamma=0$,$\alpha-t \beta+\gamma=0$,and $\alpha+\beta-t \gamma=0$. Match each entry in List-$I$ to the correct entry in List-$II$ and choose the correct option.

List-$I$	List-$II$
$(P)$ $|\vec{v}|^2$ is equal to	$(1)$ $0$
$(Q)$ If $\alpha=\sqrt{3}$,then $\gamma^2$ is equal to	$(2)$ $1$
$(R)$ If $\alpha=\sqrt{3}$,then $(\beta+\gamma)^2$ is equal to	$(3)$ $2$
$(S)$ If $\alpha=\sqrt{2}$,then $t+3$ is equal to	$(4)$ $3$
	$(5)$ $5$