If the vertices of a $\triangle ABC$ are $A=(2,3,5)$,$B=(-1,3,2)$,and $C=(3,5,-2)$,then the area of the $\triangle ABC$ (in sq. units) is

Let a vector $\vec{a}$ have a magnitude $9$. Let a vector $\vec{b}$ be such that for every $(x, y) \in \mathbb{R} \times \mathbb{R} \setminus \{(0,0)\}$,the vector $(x \vec{a} + y \vec{b})$ is perpendicular to the vector $(6y \vec{a} - 18x \vec{b})$. Then the value of $|\vec{a} \times \vec{b}|$ is equal to:

The acute angle $\theta$ between the vector $\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k}$ and the plane containing the vectors $\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + 2\hat{k}$ is:

$A$ unit vector coplanar with $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ and perpendicular to $\vec{c} = \hat{i} + \hat{j} + \hat{k}$ is

The area of a parallelogram whose two adjacent sides are represented by the vectors $\vec{a} = 3i - k$ and $\vec{b} = i + 2j$ is

For any vector $x$, where $\hat{i}, \hat{j}, \hat{k}$ have their usual meanings, the value of $(x \times \hat{i})^{2} + (x \times \hat{j})^{2} + (x \times \hat{k})^{2}$ is equal to