If $\alpha, \beta, \gamma$ are real numbers such that $(\frac{7}{3}+\beta) \hat{i}-\hat{j}+(\alpha+\gamma) \hat{k}=\frac{5}{3}(\alpha \hat{i}+\hat{j}-\hat{k})+\beta(2 \hat{j}+\hat{k})+(\hat{i}+\gamma \hat{j}+3 \hat{k})$, then $5 \alpha-9 \beta+13 \gamma=$

In a quadrilateral $ABCD$,$M$ and $N$ are the mid-points of the sides $AB$ and $CD$ respectively. If $\vec{AD} + \vec{BC} = t \vec{MN}$,then $t =$

$ABCD$ is a tetrahedron. $\bar{i}-2\bar{j}+3\bar{k}$, $-2\bar{i}+\bar{j}+3\bar{k}$, and $3\bar{i}+2\bar{j}-\bar{k}$ are the position vectors of the points $A, B, C$ respectively. $-\bar{i}+2\bar{j}-3\bar{k}$ is the position vector of the centroid of the triangular face $BCD$. If $G$ is the centroid of the tetrahedron, then $GD=$

$a$ and $b$ are non-collinear vectors. If $c=(x-2)a+b$ and $d=(2x+1)a-b$ are collinear vectors,then the value of $x = \ldots$

The value of $x$ if $x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector is:

If $\bar{a}, \bar{b}, \bar{c}$ are three non-zero vectors,no two of them are collinear,$\bar{a}+2 \bar{b}$ is collinear with $\bar{c}$,and $\bar{b}+3 \bar{c}$ is collinear with $\bar{a}$,then $\bar{a}+2 \bar{b}$ is equal to: