If $\vec{a}=(p, -2, 5)$ and $\vec{b}=(1, q, -3)$ are collinear vectors,then:

Write two different vectors having the same direction.

The unit vector in the direction of $\vec{x} = (2, 3, \sqrt{3})$ is . . . . . . .

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and the points represented by position vectors $\bar{a}-2 \bar{b}+3 \bar{c}$, $-4 \bar{a}+5 \bar{b}-6 \bar{c}$, and $x \bar{a}-9 \bar{b}+z \bar{c}$ are collinear, then $2x-z=$

$ABCD$ 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 $GD=$

Vector $\vec{x}$ is a vector in the direction of $(2, -2, 1)$ and has a magnitude of $6$ units. Vector $\vec{y}$ is a vector in the direction of $(1, 1, -1)$ and has a magnitude of $\sqrt{3}$ units. Then,$|\vec{x} + 2\vec{y}| = $ . . . . . . .