The area of a parallelogram whose adjacent sides are given by the vectors $i + 2j + 3k$ and $-3i - 2j + k$ (in square units) is

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 3\hat{i} - \hat{j} + 5\hat{k}$, and $\vec{c} = \hat{i} - 4\hat{j} - 2\hat{k}$ be three vectors. Let $\vec{r}$ be a vector perpendicular to both $\vec{b}$ and $\vec{c}$, and $\vec{r} \cdot \vec{a} = 11$. Then the vector among the following that is perpendicular to $\vec{r}$ is:

The area of the triangle,whose vertices are $A \equiv(1,-1,2)$,$B \equiv(2,1,-1)$ and $C \equiv(3,-1,2)$,is

The vectors are $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+\hat{j}$. If $\bar{c}$ is a vector such that $\bar{a} \cdot \bar{c}=|\bar{c}|$ and $|\bar{c}-\bar{a}|=2 \sqrt{2}$,and the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$ is $\frac{\pi}{4}$,then find the value of $|(\bar{a} \times \bar{b}) \times \bar{c}|$.

If $\bar{a} = \bar{i} - 2\bar{j} - 2\bar{k}$ and $\bar{b} = 2\bar{i} + \bar{j} + 2\bar{k}$ are two vectors, then $(\bar{a} + 2\bar{b}) \times (3\bar{a} - \bar{b}) = $

The unit vector perpendicular to each of the vectors $\bar{a}+\bar{b}$ and $\bar{a}-\bar{b}$,where $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ is