In a quadrilateral $PQRS$,$A$ divides $SR$ in the ratio $1:3$ and $B$ is the mid-point of $PR$. If $3SR - QR - 3PS - PQ = kAB$,then $k=$

Let $A, B, C$ be three points whose position vectors are $\overrightarrow{a} = \hat{i} + 4\hat{j} + 3\hat{k}$,$\overrightarrow{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$ (where $\alpha \in R$),and $\overrightarrow{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$. If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear,then the length of the median in $\triangle ABC$ through $A$ is:

If $a$ and $b$ are two non-collinear vectors and the vector $a+b$ bisects the angle between $a$ and $b$,then

If $\hat{i}+2 \hat{j}+3 \hat{k}$ and $3 \hat{i}+2 \hat{j}+\hat{k}$ are sides of a parallelogram,then a unit vector parallel to one of the diagonals of the parallelogram is

The ratio in which $\hat{i}+2 \hat{j}+3 \hat{k}$ divides the join of $-2 \hat{i}+3 \hat{j}+5 \hat{k}$ and $7 \hat{i}-\hat{k}$ is

The position vectors of three points $A, B$ and $C$ are $(1, 3, x), (3, 5, 8)$ and $(y, -1, -6)$ respectively. If $A, B$ and $C$ are collinear,then $(x, y) =$