Let $\vec{\alpha}=4 \hat{i}+3 \hat{j}+5 \hat{k}$ and $\vec{\beta}=\hat{i}+2 \hat{j}-4 \hat{k}$. Let $\vec{\beta}_1$ be parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ be perpendicular to $\vec{\alpha}$. If $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$,then the value of $5 \vec{\beta}_2 \cdot(\hat{i}+\hat{j}+\hat{k})$ is

Let $\bar{a}, \bar{b}, \bar{c}$ be three vectors such that $\bar{a}+\bar{b}+\bar{c}=\bar{0}$,$|\bar{a}|=3$,$|\bar{b}|=4$,and $|\bar{c}|=5$. Then,find the value of $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}$.

$7 \bar{i}-4 \bar{j}+7 \bar{k}, \bar{i}-6 \bar{j}+10 \bar{k}, -\bar{i}-3 \bar{j}+4 \bar{k}, 5 \bar{i}-\bar{j}+\bar{k}$ are the position vectors of the points $A, B, C, D$ respectively. If $p \bar{i}+q \bar{j}+r \bar{k}$ is the position vector of the point of intersection of the diagonals of the quadrilateral $ABCD$, then $p+q+r=$

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:

If $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$,then $a$ is equal to

$A$ force of magnitude $6$ acts along the vector $(9, 6, -2)$ and passes through a point $A(4, -1, -7)$. The moment of the force about the point $O(1, -3, 2)$ is