If $a = 3i - 2j + k$,$b = 2i - 4j - 3k$,and $c = -i + 2j + 2k$,then $a + b + c = \dots$

The vertices of triangle $ABC$ are $A \equiv (3, 0, 0)$,$B \equiv (0, 0, 4)$,and $C \equiv (0, 5, 4)$. Find the position vector of the point $D$ where the angle bisector of $\angle A$ meets $BC$.

The position vectors of $A$ and $B$ are $2i - 9j - 4k$ and $6i - 3j + 8k$ respectively. Then the magnitude of $\overrightarrow{AB}$ 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 $x =$

If the position vectors of the points $A, B, C, D$ given by $\hat{i}+2 \hat{j}+3 \hat{k}, 2 \hat{i}-\hat{j}+2 \hat{k}$, $\frac{1}{4}(7 \hat{i}+15 \hat{j}+15 \hat{k})$ and $\frac{1}{3}[7 \hat{i}+2 \hat{j}+(5+3 a) \hat{k}]$ respectively are such that $|AC|=|BD|$, then $16(3a-1)^2=$

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors. Then the point of intersection of the line joining the points $\vec{a}+\vec{b}+\vec{c}, \vec{a}-\vec{b}+3 \vec{c}$ and the line joining the points $2 \vec{a}-\vec{b}+\vec{c}, \vec{a}-2 \vec{b}+4 \vec{c}$ is