$A$ vector perpendicular to both of the vectors $i + j + k$ and $i + j$ is

The area of the triangle whose vertices are $A(1, 1, 1)$,$B(1, 2, 3)$,and $C(2, 3, 1)$ is . . . . . . .

Let $\overrightarrow{a}=3 \hat{i}+2 \hat{j}+\hat{k}$,$\overrightarrow{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\overrightarrow{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$ and $(\overrightarrow{a}-\overrightarrow{b}+\hat{i}) \cdot \overrightarrow{c}=-3$. Then $|\overrightarrow{c}|^2$ is equal to . . . . . . .

For two given vectors $\bar{a}$ and $\bar{b}$,if the vectors $\overline{A}$ and $\overline{B}$ are such that $\overline{A}+\overline{B}=\bar{a}$,$\overline{A} \times \overline{B}=\bar{b}$,and $\overline{A} \cdot \bar{a}=1$,then $\overline{A}=$

If $|\bar{u}| = 8$ and $|\bar{v}| = 12$ with an angle of $150^{\circ}$ between them,then find $|\bar{u} \times \bar{v}|$.

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: