Find $|\vec{x}|$,if for a unit vector $\vec{a}$,$(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 12$.

$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $

If $\hat{i}+\hat{j}+\hat{k}, 2 \hat{i}+5 \hat{j}, 3 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\hat{i}-6 \hat{j}-\hat{k}$ are the position vectors of points $A, B, C$ and $D$ respectively,then find the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$. Deduce that $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are collinear.

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$,then the position vector of the point in which the bisector of $\angle B$ meets $CA$ is:

If the position vectors of the vertices $A, B$ and $C$ of a $\Delta ABC$ are respectively $4\hat{i} + 7\hat{j} + 8\hat{k}$,$2\hat{i} + 3\hat{j} + 4\hat{k}$ and $2\hat{i} + 5\hat{j} + 7\hat{k}$,then the position vector of the point,where the bisector of $\angle A$ meets $BC$ is

If $ 2 \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| $,then the angle between $ \vec{a} $ and $ \vec{b} $ is: (in $^{\circ}$)