If $A(1,-1,2)$,$B(5,7,-6)$,$C(3,4,-10)$,and $D(-1,-4,-2)$ are the vertices of a quadrilateral $ABCD$,then its area is:

Let $\vec{a}$ be a unit vector and $\vec{b}$ be a nonzero vector not parallel to $\vec{a}$. The angles of the triangle,two of whose sides are represented by $\sqrt{3}(\vec{a} \times \vec{b})$ and $\vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$,are

If the vector $\vec{b} = 3\hat{j} + 4\hat{k}$ is written as the sum of a vector $\vec{b_1}$,parallel to $\vec{a} = \hat{i} + \hat{j}$ and a vector $\vec{b_2}$,perpendicular to $\vec{a}$,then $\vec{b_1} \times \vec{b_2}$ is equal to

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$,$\overrightarrow{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$,$\overrightarrow{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\overrightarrow{d}$ be a vector such that $\overrightarrow{b} \times \overrightarrow{d} = \overrightarrow{c} \times \overrightarrow{d}$ and $\overrightarrow{a} \cdot \overrightarrow{d} = 4$. Then $|(\overrightarrow{a} \times \overrightarrow{d})|^2$ is equal to . . . . . . .

Find the area of a triangle having the points $A(1, 1, 1)$,$B(1, 2, 3)$,and $C(2, 3, 1)$ as its vertices.

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: