If $|\vec{a}| = 4$,$|\vec{b}| = 2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then $|\vec{a} \times \vec{b}|^2 = \dots$

If $\overrightarrow{a}=2 \hat{i}-5 \hat{j}+8 \hat{k}$ and $\overrightarrow{b}=7 \hat{i}-5 \hat{j}+3 \hat{k}$ are two vectors and $(2 \overrightarrow{a}-3 \overrightarrow{b}) \times(4 \overrightarrow{a}+\overrightarrow{b})=x \hat{i}+y \hat{j}+z \hat{k}$,then $x+y+z=$

If $a, b$ and $c$ are three vectors such that $a \times b = c$,$b \times c = a$ and $a, b, c$ are mutually perpendicular to each other,then $|b|$ is equal to

$A$ force $\overrightarrow{F} = 2i + j - k$ acts at a point $A$,whose position vector is $2i - j$. The moment of $\overrightarrow{F}$ about the origin is:

Find a vector of magnitude $6,$ which is perpendicular to both the vectors $\vec{a} = 2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b} = 4 \hat{i}-\hat{j}+3 \hat{k}$.

If $a = (1, -1, 1)$ and $c = (-1, -1, 0),$ then the vector $b$ satisfying $a \times b = c$ and $a \cdot b = 1$ is