If the points with position vectors $\alpha \hat{i} + 10 \hat{j} + 13 \hat{k}$,$6 \hat{i} + 11 \hat{j} + 11 \hat{k}$,and $\frac{9}{2} \hat{i} + \beta \hat{j} - 8 \hat{k}$ are collinear,then $(19 \alpha - 6 \beta)^2$ is equal to $...........$.

Suppose $ABCDE$ is a pentagon. The resultant vector of the vectors $\vec{AB}, \vec{AE}, \vec{BC}, \vec{DC}, \vec{ED}$ and $\vec{AC}$ is

$\bar{a}$ and $\bar{b}$ are non-collinear vectors. If $\bar{p} = (2x + 1)\bar{a} - \bar{b}$ and $\bar{q} = (x - 2)\bar{a} + \bar{b}$ are collinear vectors,then $x =$

The magnitude of the projection of the vector $\vec{a} = -\hat{i} + 2\hat{j} - \hat{k}$ on the unit vector $\hat{i}$ is . . . . . . .

If the position vector of a point $A$ is $a + 2b$ and a point $P$ divides $AB$ in the ratio $2:3$,where the position vector of $P$ is $a$,then the position vector of $B$ is:

Vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ are of the same length and they make equal angles with each other when taken in pairs. If $\vec{a} = \hat{i} - \hat{j}$,$\vec{b} = \hat{j} + \hat{k}$,and $\vec{c}$ makes an obtuse angle with the $x$-axis,find the vector $\vec{c}$.