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

$a=3 \hat{i}+\hat{j}-\hat{k}, b=\hat{i}-4 \hat{j}+5 \hat{k}, c=4 \hat{i}+5 \hat{j}-\hat{k}$ are three vectors and a vector $r$ is perpendicular to both the vectors $b$ and $c$. If $r \cdot a=9$,then $r=$

Let $\vec{a} = \vec{j} - \vec{k}$ and $\vec{c} = \vec{i} - \vec{j} - \vec{k}$. Find the vector $\vec{b}$ satisfying $\vec{a} \times \vec{b} + \vec{c} = 0$ and $\vec{a} \cdot \vec{b} = 3$.

Show that $(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})$.

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 $\frac{\pi}{6}$,then $|(\overline{a} \times \overline{b}) \times \overline{c}|$ is

The area of a parallelogram whose diagonals are the vectors $2 \bar{a}-\bar{b}$ and $4 \bar{a}-5 \bar{b}$,where $\bar{a}$ and $\bar{b}$ are unit vectors forming an angle of $45^{\circ}$ is