The area of the triangle having vertices as $i - 2j + 3k,$ $- 2i + 3j - k,$ and $4i - 7j + 7k$ is

The area of the parallelogram for which the vectors $\hat{i}+\hat{j}+2 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$ are adjacent sides is equal to

Let $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\vec{a} \times \vec{b}| = r$ is possible if

Let $\bar{a}$,$\bar{b}$,and $\bar{c}$ be unit vectors. Suppose that $\bar{a} \cdot \bar{b} = \bar{a} \cdot \bar{c} = 0$ and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{6}$. Then $\bar{a}$ is equal to:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{j}-\hat{k}$ are given vectors,then a vector $\vec{b}$ satisfying the equations $\vec{a} \times \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=3$ is

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 . . . . . . .