The points with position vectors $10\,i + 3\,j$,$12\,i - 5\,j$,and $a\,i + 11\,j$ are collinear,if $a = $

If the vectors $3\,i + 2\,j - k$ and $6\,i - 4xj + yk$ are parallel,then the value of $x$ and $y$ will be

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 vectors $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a $\triangle ABC$. The length of the median through $A$ is:

If the points whose position vectors are $2 \hat{i}+\hat{j}+\hat{k}$,$6 \hat{i}-\hat{j}+2 \hat{k}$,and $14 \hat{i}-5 \hat{j}+p \hat{k}$ are collinear,then the value of $p$ is:

If $\hat{i}$ is the position vector of the centroid $G$ of triangle $ABC$ and $2\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}-4\hat{k}$ are respectively the position vectors of its vertices $A$ and $B$,then $AG^2+BG^2+CG^2=$