If $|\overrightarrow{a}|=2, |\overrightarrow{b}|=3$ and $\overrightarrow{a}, \overrightarrow{b}$ are mutually perpendicular,then the area of the triangle whose vertices are $\overrightarrow{0}, \overrightarrow{a}+\overrightarrow{b}, \overrightarrow{a}-\overrightarrow{b}$ is

The unit vectors perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ are:

$A$ vector which is orthogonal to the vector $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and coplanar with the vectors $\vec{b} = 3\hat{i} + 2\hat{j}$ and $\vec{c} = 2\hat{i} + \hat{j} + 3\hat{k}$ is

Let $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+\hat{j}$. If $\bar{c}$ is a vector such that $\bar{a} \cdot \bar{c}=|\bar{c}|$,$|\bar{c}-\bar{a}|=2 \sqrt{2}$ and the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$ is $60^{\circ}$. Then $|(\bar{a} \times \bar{b}) \times \bar{c}|=$

If $\overline{a} = \hat{i} + \hat{j} + \hat{k}$ and $\overline{b} = \hat{j} - \hat{k}$,then the vector $\overline{r}$ satisfying $\overline{a} \times \overline{r} = \overline{b}$ and $\overline{a} \cdot \overline{r} = 3$ is

The area of the triangle with vertices $(1,2,0)$,$(1,0,2)$ and $(0,3,1)$ is.