If $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$,then under what condition are the three points $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ collinear?

The position vector of a point at a distance of $3\sqrt{11}$ units from $i - j + 2k$ on a line passing through the points $i - j + 2k$ and $3i + j + k$ is

If the points $P, Q$ and $R$ have the position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$,$-2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $-8 \hat{i}+13 \hat{j}$ respectively,then these points are

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}$ be two vectors such that $\vec{a} \cdot \vec{b}=1$, $\cos(\theta) = \frac{1}{3}$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, and the components of $\vec{b}$ with respect to $(\hat{i}, \hat{j}, \hat{k})$ are integers. Then the number of possible vectors that represent $\vec{b}$ is

$A$ vector $r$ is equally inclined with the coordinate axes. If the tip of $r$ is in the positive octant and $|r| = 6$,then $r$ is

$A, B, C, D$ are any four points. If $E$ and $F$ are midpoints of $AC$ and $BD$ respectively, then $\vec{AB} + \vec{CB} + \vec{CD} + \vec{AD} =$