If $\overline{a}=\hat{i}+\hat{j}+\hat{k}$,$\overline{a} \cdot \overline{b}=1$ and $\overline{a} \times \overline{b}=\hat{j}-\hat{k}$,then $\overline{b}$ is

The position vectors of the points $A, B$ and $C$ are $i + j, j + k$ and $k + i$ respectively. The vector area of the $\Delta ABC = \pm \frac{1}{2} \vec{\alpha}$ where $\vec{\alpha} = $

Let the vectors $\overline{a}, \overline{b}, \overline{c}$ and $\overline{d}$ be such that $(\overline{a} \times \overline{b}) \times(\overline{c} \times \overline{d})=\overline{0}$. Let $P_1$ and $P_2$ be the planes determined by the pair of vectors $\overline{a}, \overline{b}$ and $\overline{c}, \overline{d}$ respectively,then the angle between $P_1$ and $P_2$ is

If $\theta$ is the angle between $\vec{f}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{g}=2 \hat{i}-3 \hat{j}+a \hat{k}$ and $\sin \theta=\sqrt{\frac{24}{28}}$,then $7 a^2+24 a=$

$A (2,6,2), B (-4,0, \lambda), C (2,3,-1)$ and $D (4,5,0)$,with $|\lambda| \leq 5$,are the vertices of a quadrilateral $ABCD$. If its area is $18$ square units,then $5-6 \lambda$ is equal to $.........$.

Let $\overrightarrow{OA}=2 \overrightarrow{a}$,$\overrightarrow{OB}=6 \overrightarrow{a}+5 \overrightarrow{b}$ and $\overrightarrow{OC}=3 \overrightarrow{b}$,where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is $15$ sq. units,then the area (in sq. units) of the quadrilateral $OABC$ is equal to :