If three points $A, B, C$ are collinear,whose position vectors are $i - 2j - 8k$,$5i - 2k$,and $11i + 3j + 7k$ respectively,then the ratio in which $B$ divides $AC$ is

Let $\vec{\alpha} = (\lambda - 2) \vec{a} + \vec{b}$ and $\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$ be two given vectors where $\vec{a}$ and $\vec{b}$ are non-collinear. The value of $\lambda$ for which vectors $\vec{\alpha}$ and $\vec{\beta}$ are collinear is:

If $D, E$ and $F$ are respectively mid-points of $AB, AC$ and $BC$ in $\triangle ABC$,then $\overrightarrow{BE} + \overrightarrow{AF}$ is equal to

In $\Delta ABC$,if $A(2, 4, -1)$,$B(4, 5, 1)$,and $C(3, 6, -3)$,then $\Delta ABC$ is a ... triangle.

The number of unit vectors of the form $a \hat{i} + b \hat{j} + c \hat{k}$,where $a, b, c \in W$ is

$3\overrightarrow{OD} + \overrightarrow{DA} + \overrightarrow{DB} + \overrightarrow{DC} = $