Three forces $i + 2j - 3k$,$2i + 3j + 4k$,and $i - j + k$ are acting on a particle at the point $(0, 1, 2)$. The magnitude of the moment of the forces about the point $(1, -2, 0)$ is

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $2 \vec{a}+3 \vec{b}+4 \vec{c}=\vec{0}$. Then $|\vec{b} \times \vec{c}|=$

If $4i + 7j + 8k$,$2i + 3j + 4k$ and $2i + 5j + 7k$ are the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$. The position vector of the point where the bisector of angle $A$ meets $BC$ is

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\overrightarrow{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\overrightarrow{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}$ be three vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{b} = \overrightarrow{c} \times \overrightarrow{b}$ and $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,then $25|\overrightarrow{r}|^2$ is equal to:

The angle between two vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ is . . . . . . .

The value of $\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})$ is