If $C$ is the midpoint of $AB$ and $P$ is any point outside $AB$,then

Let $\overline{i}-2 \overline{j}+\overline{k}, \overline{i}+\overline{j}-2 \overline{k}, 2 \overline{i}-\overline{j}-\overline{k}$ and $\overline{i}+\overline{j}+\overline{k}$ be the position vectors of four points $A, B, C$ and $D$ respectively. If a point $P$ divides $AB$ in the ratio $2:1$ internally and a point $Q$ divides $CD$ in the ratio $1:2$ externally,then the ratio in which the point with position vector $5 \overline{i}-6 \overline{j}-5 \overline{k}$ divides $PQ$ is

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 $BC$.

If the collinear points $A, B$ and $C$ have position vectors respectively $(1, x, 3), (3, 4, 7)$ and $(y, -2, -5)$,then $x+y=$

$(a \cdot i)i + (a \cdot j)j + (a \cdot k)k = $

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 $16(3a-1)^2=$