Find $|a \times b|^2$,if $|a|=2, |b|=3$ and the angle between $a$ and $b$ is $\theta = \frac{\pi}{6}$.

Let $a = 2i + j - 2k$ and $b = i + j$. If $c$ is a vector such that $a \cdot c = |c|$,$|c - a| = 2\sqrt{2}$,and the angle between $(a \times b)$ and $c$ is $30^\circ$,then $|(a \times b) \times c| = $

$A$ unit vector perpendicular to both $i+j+k$ and $2i+j+3k$ is

Let $\overline{a}=\alpha \hat{i}+3 \hat{j}-\hat{k}$,$\overline{b}=3 \hat{i}-\beta \hat{j}+4 \hat{k}$ and $\overline{c}=\hat{i}+2 \hat{j}-2 \hat{k}$,where $\alpha, \beta \in R$,be three vectors. If the projection of $\overline{a}$ on $\overline{c}$ is $\frac{10}{3}$ and $\overline{b} \times \overline{c}=-6 \hat{i}+10 \hat{j}+7 \hat{k}$,then the value of $\alpha^2+\beta^2-\alpha \beta$ is equal to

Let $\vec{a}=6 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}$. If $\vec{c}$ is a vector such that $|\vec{c}| \geq 6$,$\vec{a} \cdot \vec{c}=6|\vec{c}|$,$|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^{\circ}$,then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to:

Given $a = i + j - k$,$b = -i + 2j + k$,and $c = -i + 2j - k$. $A$ unit vector perpendicular to both $a + b$ and $b + c$ is