The adjacent sides of a parallelogram are $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$. Find its area.

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}+\hat{j}-2\hat{k}, \vec{c}=\hat{i}-2\hat{j}+3\hat{k}$ and $\vec{d}=-4\hat{i}+5\hat{j}-3\hat{k}$. If $\vec{d}=x(\vec{b} \times \vec{c})-\frac{7}{9}(\vec{c} \times \vec{a})+z(\vec{a} \times \vec{b})$,then find the value of $x$.

If $C$ is a given non-zero scalar and $\overline{A}$ and $\overline{B}$ are given non-zero vectors such that $\overline{A}$ is perpendicular to $\overline{B}$. If vector $\overline{X}$ is such that $\overline{A} \cdot \overline{X} = C$ and $\overline{A} \times \overline{X} = \overline{B}$,then $\overline{X}$ is given by:

Let $F=2 \hat{i}+2 \hat{j}+5 \hat{k}$,$A=(1,2,5)$,$B=(-1,-2,-3)$ and $BA \times F=4 \hat{i}+6 \hat{j}+2 \lambda \hat{k}$,then $\lambda=$

$A$ unit vector orthogonal to the vector $\vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k}$ and coplanar with the vectors $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + \hat{k}$ is:

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