If the vectors $\overline{AB}=3 \hat{i}+4 \hat{k}$ and $\overline{AC}=5 \hat{i}-2 \hat{j}+4 \hat{k}$ are the sides of the triangle $ABC$,then the length of the median through $A$ is:

The position vectors of $A$ and $B$ are $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 3\hat{k}$. The position vector of the midpoint of the line segment $AB$ is:

If $\vec{r} = 3\hat{i} + 2\hat{j} - 5\hat{k}$,$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$,and $\vec{c} = -2\hat{i} + \hat{j} - 3\hat{k}$ such that $\vec{r} = \lambda\vec{a} + \mu\vec{b} + \gamma\vec{c}$,then -

The points with position vectors $60\,i + 3\,j$,$40\,i - 8\,j$,and $a\,i - 52\,j$ are collinear,if $a = $

In $\triangle PQR$,$(4 \hat{i}+3 \hat{j}+6 \hat{k})$,$(2 \hat{i}+2 \hat{j}+3 \hat{k})$ and $(3 \hat{i}+\hat{j}+3 \hat{k})$ are the position vectors of the vertices $P, Q$ and $R$ respectively. Then the position vector of the point of intersection of the angle bisector of $P$ with $QR$ is

Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2 \vec{a}+\vec{b})$ and $(\vec{a}-3 \vec{b})$ externally in the ratio $1:2$. Also,show that $P$ is the mid-point of the line segment $RQ$.