The unit vector perpendicular to both the vectors $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ is

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices of a triangle,show that $\frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}]$ gives the vector area of the triangle. Hence,deduce the condition that the three points $\vec{a}, \vec{b}, \vec{c}$ are collinear. Also,find the unit vector normal to the plane of the triangle.

$\text{If } \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k} \text{ and } \vec{c} = \hat{i} - \hat{j} \text{ and if } 6\hat{i} + 2\hat{j} + 3\hat{k} = \lambda_1(\vec{a} \times \vec{b}) + \lambda_2(\vec{b} \times \vec{c}) + \lambda_3(\vec{c} \times \vec{a}), \text{ then } (\lambda_1, \lambda_2, \lambda_3) = $

Let $\overrightarrow{OA}=\overrightarrow{a}$,$\overrightarrow{OB}=12 \overrightarrow{a}+4 \overrightarrow{b}$,and $\overrightarrow{OC}=\overrightarrow{b}$,where $O$ is the origin. If $S$ is the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$,then the ratio of the area of the quadrilateral $OABC$ to the area of $S$ is:

If the area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is $20$ square units,then the area of the parallelogram having $3 \bar{a} + \bar{b}$ and $2 \bar{a} + 3 \bar{b}$ as two adjacent sides in square units is

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar unit vectors such that $\bar{a} \times(\bar{b} \times \bar{c})=\frac{(\bar{b}+\bar{c})}{\sqrt{2}}$,then the angle between $\bar{a}$ and $\bar{b}$ is