The position vectors of the points $A, B, C$ are $(2i + j - k)$,$(3i - 2j + k)$,and $(i + 4j - 3k)$ respectively. These points

$A$ 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

$(3, 0, 2)$ and $(0, 2, k)$ are the direction ratios of two lines and $\theta$ is the angle between them. If $|\cos \theta| = \frac{6}{13}$,then $k =$

If the position vectors of points $P, Q, R,$ and $S$ are $2\hat{i} + 3\hat{j} + 5\hat{k}$,$\hat{i} + 2\hat{j} + 3\hat{k}$,$-5\hat{i} + 4\hat{j} - 2\hat{k}$,and $\hat{i} + 10\hat{j} + 10\hat{k}$ respectively,then:

Let $D, E, F$ be the midpoints of the sides $BC, CA, AB$ of a triangle $ABC$ respectively. Then $\vec{AD} + \vec{BE} + \vec{CF} = \dots$

If $\bar{a}, \bar{b}, \bar{c}$ are non-zero vectors such that no two of them are parallel,and $\bar{a} + \bar{b}$ is parallel to $\bar{c}$,and $\bar{b} + \bar{c}$ is parallel to $\bar{a}$,then $\bar{a} + \bar{b} + \bar{c} = $