$(\bar{a}+2 \bar{b}-\bar{c}) \cdot \{(\bar{a}-\bar{b}) \times (\bar{a}-\bar{b}-\bar{c})\} = $

If the origin $O(0,0,0)$ and the points $P(2,3,4)$,$Q(1,2,3)$,and $R(x, y, z)$ are co-planar,then:

If $\vec{u} = \hat{i} + 2\hat{j} - 2\hat{k}$, $\vec{v} = 2\hat{i} + \hat{k}$ and $\vec{w}$ is a unit vector, then the maximum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is

Unit vectors $a, b$ and $c$ are coplanar. $A$ unit vector $d$ is perpendicular to them. 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 $c$ is:

If $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$,$\vec{b}=\hat{i}+3 \hat{j}-\hat{k}$ and $\vec{c}=3 \hat{i}-\hat{j}-2 \hat{k}$,then the value of $\left|\begin{array}{lll}\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c}\end{array}\right|$ is:

The volume of the tetrahedron formed by the vectors $\vec{a}, \vec{b}, \vec{c}$ is $3$. Then the volume of the parallelepiped formed by the coterminous edges $\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}$ is: