The vector $x\hat{i} + y\hat{j} + z\hat{k}$ makes an acute angle $\cot^{-1} \sqrt{2}$ with the plane containing the vectors $(2, 3, -1)$ and $(1, -1, 2)$. Then,

Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be the position vectors of the vertices $A, B, C$ respectively of $\triangle ABC$. The vector area of $\triangle ABC$ is:

The area of the parallelogram whose diagonals are the vectors $2\vec{a} - \vec{b}$ and $4\vec{a} - 5\vec{b},$ where $\vec{a}$ and $\vec{b}$ are unit vectors forming an angle of $45^{\circ},$ 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:

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.

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $\overline{a} \neq \overline{0}$ and $\overline{a} \times \overline{b} = 2 \overline{a} \times \overline{c}$,$|\overline{a}| = |\overline{c}| = 1$,$|\overline{b}| = 4$ and $|\overline{b} \times \overline{c}| = \sqrt{15}$. If $\overline{b} - 2 \overline{c} = \lambda \overline{a}$,then $\lambda$ is