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 $\lambda=$

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. If $\overline{c}$ is a vector such that $\overline{a} \cdot \overline{c}=|\overline{c}|$,$|\overline{c}-\overline{a}|=2 \sqrt{2}$,and the angle between $\overline{a} \times \overline{b}$ and $\overline{c}$ is $\frac{2 \pi}{3}$,then $|(\overline{a} \times \overline{b}) \times \overline{c}|=$

The area of the quadrilateral $ABCD$ with vertices $A(2, 1, 1)$,$B(1, 2, 5)$,$C(-2, -3, 5)$,and $D(1, -6, -7)$ is equal to

The vector $x$ is perpendicular to the vectors $a=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $b=18 \hat{i}-22 \hat{j}-5 \hat{k}$,and it makes an obtuse angle with $\hat{j}$. If $|x|=14$,then $x=$

If $\overline{a}=\hat{i}+\hat{j}+\hat{k}$,$\overline{a} \cdot \overline{b}=1$ and $\overline{a} \times \overline{b}=\hat{j}-\hat{k}$,then $\overline{b}$ is

Let $\overrightarrow{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\hat{i}-2 \hat{j}-3 \hat{k}$ be two vectors. If $A_1$ is the area of the quadrilateral having $\vec{a}, \vec{b}$ as its diagonals and $A_2$ is the area of the parallelogram having $\overrightarrow{a}, \overrightarrow{b}$ as its two adjacent sides,then $A_1 \cdot A_2=$