The position vectors of the vertices $A, B$ and $C$ of a triangle are $2 \hat{i}-3 \hat{j}+3 \hat{k}$,$2 \hat{i}+2 \hat{j}+3 \hat{k}$ and $-\hat{i}+\hat{j}+3 \hat{k}$ respectively. Let $l$ denote the length of the angle bisector $AD$ of $\angle BAC$,where $D$ is on the line segment $BC$. Then $2 l^2$ equals:

If $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$,$\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ and the length of the internal bisector of $\angle BOA$ of triangle $AOB$ is $k$,then $9k^2 =$

$a$ and $b$ are non-collinear vectors,$|a|=2 \sqrt{2}$,$|b|=3$ and the angle between $a$ and $b$ is $45^{\circ}$. Then,the lengths of the diagonals of the parallelogram whose adjacent sides are represented by the vectors $5a+2b$ and $a-3b$ are

If $\bar{a}, \bar{b}$ and $\bar{c}$ are vectors such that $|\bar{a}| = |\frac{\bar{b}}{2}| = |\frac{\bar{c}}{3}| = 1$; $\bar{b}$ and $\bar{c}$ are perpendicular; and the projections of $\bar{b}$ and $\bar{c}$ on $\bar{a}$ are equal,then $|\bar{a} - \bar{b} + \bar{c}| = $

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\overrightarrow{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$. If $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{r}$,$\overrightarrow{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$ and $\overrightarrow{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$,where $\alpha \in R$,then the value of $\alpha + |\overrightarrow{r}|^{2}$ is equal to:

$A$ vector $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k}$ (where $\alpha, \beta \in R$) lies in the plane of the vectors $\overrightarrow{b} = \hat{i} + \hat{j}$ and $\overrightarrow{c} = \hat{i} - \hat{j} + 4 \hat{k}$. If $\overrightarrow{a}$ bisects the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$,then: