In $\triangle ABC$,if $D$ and $E$ are the mid-points of the sides $BC$ and $CA$ respectively,then $2(\vec{AD}+\vec{EB})=$

The position vectors of three points $A, B$ and $C$ are $(1, 3, x), (3, 5, 8)$ and $(y, -1, -6)$ respectively. If $A, B$ and $C$ are collinear,then $(x, y) =$

If $\vec{a}$ is any vector in space,then

If the points with position vectors $10\hat{i} + 3\hat{j}$,$12\hat{i} - 5\hat{j}$,and $a\hat{i} + 11\hat{j}$ are collinear,then the value of $a$ is:

If $P \hat{i}-2 \hat{j}+3 \hat{k}$,$2 \hat{i}+3 \hat{j}-4 \hat{k}$,and $4 \hat{i}+13 \hat{j}-18 \hat{k}$ are the position vectors of three collinear points $A$,$B$,and $C$ respectively,then the vector in the direction of $AB$ of length $|P|$ units is

$ABCDEF$ is a regular hexagon. Find the sum of the vectors $\vec{BE} + \vec{BC} + \vec{EF} + \vec{BA} + \vec{CF} + \vec{AF}$.