Let $v = 2i + j - k$ and $w = i + 3k$. If $u$ is any unit vector, then the maximum value of the scalar triple product $[u v w]$ is

If $\bar{a} = \hat{i} - \hat{j}$,$\bar{b} = \hat{j} - \hat{k}$,and $\bar{c} = \hat{k} - \hat{i}$,then a unit vector $\bar{d}$ such that $\bar{a} \cdot \bar{d} = 0$ and $[\bar{b} \bar{c} \bar{d}] = 0$ is:

If $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar vectors and $p, q$ are real numbers,then the equality $[3\vec{u}, p\vec{v}, p\vec{w}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w}, q\vec{v}, q\vec{u}] = 0$ holds for:

The volume of the parallelepiped whose coterminous edges are $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = 3\hat{i} - \hat{j} - \hat{k}$, and $\vec{c} = \hat{j} + 3\hat{k}$ is:

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and the points $P_1 = \lambda \vec{a}+3 \vec{b}-\vec{c}$, $P_2 = \vec{a}-\lambda \vec{b}+3 \vec{c}$, $P_3 = 3 \vec{a}+4 \vec{b}-\lambda \vec{c}$, and $P_4 = \vec{a}-6 \vec{b}+6 \vec{c}$ are coplanar, then one of the values of $\lambda$ is

If the points $A(2,1,-1), B(0,-1,0), C(4,0,4)$ and $D(2,0,x)$ are coplanar,then $x=$