If the vectors $\hat{i}+2 \hat{j}+x \hat{k}$ and $y \hat{i}+6 \hat{j}+4 \hat{k}$ are collinear,then the values of $x$ and $y$ are respectively,

If $C$ is the mid-point of the line segment $AB$ and $P$ is any point outside the line $AB$,then

Find the unit vector in the direction of vector $\overrightarrow{PQ},$ where $P$ and $Q$ are the points $(1, 2, 3)$ and $(4, 5, 6),$ respectively.

$D, E, F$ are points on the sides $BC, CA$ and $AB$ of a $\triangle ABC$ respectively,dividing them in the ratios $2:3, 1:2, 3:1$ internally. The lines $BE$ and $CF$ intersect on the line $AD$ at $P$. If $\overrightarrow{AP} = x_1 \overrightarrow{AB} + y_1 \overrightarrow{AC}$,then $x_1 + y_1 =$

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

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