If $ABCDEF$ is a regular hexagon,and $\vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF} = k \vec{AD}$,then $k = \dots$

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If $\lambda=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ and $\vec{d}=\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}$,then the ordered pair $(\lambda, \vec{d})$ is equal to:

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

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 ............. $unit$.

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 $|v| =$.

The points $D, E, F$ divide $BC, CA$ and $AB$ of the triangle $ABC$ in the ratio $1:4, 3:2$ and $3:7$ respectively and the point $K$ divides $AB$ in the ratio $1:3$,then $(\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}) : \overrightarrow{CK}$ is equal to