$a$ is perpendicular to both $b$ and $c$. The angle between $b$ and $c$ is $\frac{2 \pi}{3}$. If $|a|=2$, $|b|=3$, and $|c|=4$, then $c \cdot (a \times b)$ is equal to (in $\sqrt{3}$)

Let $\vec{u} = a\hat{i} + b\hat{j} + c\hat{k}$,$\vec{v} = b\hat{i} + c\hat{j} + a\hat{k}$,and $\vec{w} = c\hat{i} + a\hat{j} + b\hat{k}$. If $[\vec{u} \, \vec{v} \, \vec{w}] = 0$ and $\vec{w} = \lambda \vec{x} + \mu \vec{y}$ where $(a + b + c) \neq 0$ and $\lambda, \mu \neq 0$,then the vectors $\vec{x}, \vec{y}, \vec{u}, \vec{v}, \vec{w}$ are:

If $\overline{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\overline{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$,then the value of $(2 \bar{a}-\bar{b}) \cdot [(\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b})] = $

Unit vectors $a, b, c$ are coplanar. $A$ unit vector $d$ is perpendicular to the given vectors. If $(a \times b) \times (c \times d) = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$ and the angle between $a$ and $b$ is $30^{\circ}$,then $c = ....$

Let $\vec{c}$ be a vector coplanar with the unit vectors $\vec{a}$ and $\vec{b}$, and let $\vec{d}$ be the unit vector perpendicular to $\vec{a}$, $\vec{b}$, and $\vec{c}$. If $[\vec{a} \vec{b} \vec{d}] \vec{c} - [\vec{a} \vec{b} \vec{c}] \vec{d} = \hat{i} + 2\hat{j} + 2\hat{k}$ and the angle between $\vec{a}$ and $\vec{b}$ is $30^{\circ}$, then $|\vec{c}| =$

$ [\vec{a}+2 \vec{b}-\vec{c}, \vec{a}-\vec{b}, \vec{a}-\vec{b}-\vec{c}] $