Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$,let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$. If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$,then $|\vec{c}|^2$ is equal to:

Let $\vec{\alpha}=\hat{i}+\hat{j}+\hat{k}$,$\vec{\beta}=\hat{i}-\hat{j}-\hat{k}$ and $\vec{\gamma}=-\hat{i}+\hat{j}-\hat{k}$ be three vectors. $A$ vector $\vec{\delta}$,in the plane of $\vec{\alpha}$ and $\vec{\beta}$,whose projection on $\vec{\gamma}$ is $\frac{1}{\sqrt{3}}$,is given by

The vector $2\hat{i} + a\hat{j} + \hat{k}$ is perpendicular to the vector $2\hat{i} - \hat{j} - \hat{k},$ if $a = $

$AB=a$ and $AC=b$ are the sides of $\triangle ABC$. $P$ is a point on $AB$ and $Q$ is a point on $BC$ such that $\frac{AP}{PB}=\frac{1}{2}$ and $\frac{BQ}{QC}=\frac{1}{2}$. If the point of intersection of $AQ$ and $CP$ is $D$ and the area of $\triangle BCD$ is $7$ square units,then the area of the $\triangle ABC$ (in the same square units) is

Let $\vec{a}=\hat{i}-3 \hat{j}+7 \hat{k}$,$\vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+2 \vec{b}) \times \vec{c}=3(\vec{c} \times \vec{a})$. If $\vec{a} \cdot \vec{c}=130$,then $\vec{b} \cdot \vec{c}$ is equal to ....................

Let $\sqrt{3} \hat{i} + \hat{j}$,$\hat{i} + \sqrt{3} \hat{j}$ and $\beta \hat{i} + (1 + \beta) \hat{j}$ respectively be the position vectors of the points $A, B$ and $C$ with respect to the origin $O$. If the distance of $C$ from the bisector of the acute angle between $OA$ and $OB$ is $\frac{3}{\sqrt{2}}$,then the sum of all possible values of $\beta$ is: