If points $D, E, F$ divide the sides $BC, CA, AB$ of $\triangle ABC$ in the ratios $1:4, 3:2, 3:7$ respectively,and point $K$ divides $AB$ in some ratio,then $(\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}) : \overrightarrow{CK} = ......$

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}, \overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{d}=\hat{i}-\hat{j}-\hat{k}$,then match the following List-$I$ with List-$II$:

List-$I$	List-$II$
$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b}$	$(A)$ $\overrightarrow{a} \cdot \overrightarrow{d}$
(ii) $\overrightarrow{b} \cdot \overrightarrow{c}$	$(B)$ $3$
(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$	$(C)$ $\overrightarrow{b} \cdot \overrightarrow{d}$
(iv) $\overrightarrow{b} \times \overrightarrow{c}$	$(D)$ $2\hat{i}-2\hat{k}$
	$(E)$ $2\hat{j}+2\hat{k}$
	$(F)$ $4$

Given that $a$ and $b$ are two unit non-collinear vectors,if $u = a - (a \cdot b)b$ and $v = a \times b$,then find $|v| =$.

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 $B$?

If $\vec{a}, \vec{b}$ and $\vec{c}$ are non-coplanar vectors and if $\vec{d}$ is such that $\vec{d} = \frac{1}{x}(\vec{a} + \vec{b} + \vec{c})$ and $\vec{d} = \frac{1}{y}(\vec{b} + \vec{c} + \vec{d})$ where $x$ and $y$ are non-zero real numbers,then $\frac{1}{xy}(\vec{a} + \vec{b} + \vec{c} + \vec{d})$ is equal to:

If $ABCDEF$ is a regular hexagon,then $\overrightarrow{AD} + \overrightarrow{EB} + \overrightarrow{FC} = .....$