If $\left[ {\vec a \,\vec b \,\vec c } \right] = 4$,then $\left[ {\vec a \times \vec b, \vec b \times \vec c, \vec c \times \vec a } \right] = \dots$

If $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 3\hat{k}$ and $\vec{c}$ is a unit vector, then the maximum value of the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is

If the volume of a tetrahedron whose vertices are $A \equiv (1, -6, 10)$,$B \equiv (-1, -3, 7)$,$C \equiv (5, -1, k)$,and $D \equiv (7, -4, 7)$ is $11$ cubic units,then the value of $k$ is:

If the volume of the tetrahedron whose coterminous edges are given by the vectors $\vec{a} = -2\hat{i} + 3\hat{j} - 3\hat{k}$, $\vec{b} = 4\hat{i} + 5\hat{j} + (\lambda - 10)\hat{k}$, and $\vec{c} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ is $11$ cubic units, then the sum of the possible values of $\lambda$ is

If the points $2a+3b-c, a-2b+3c, 3a+\lambda b-2c$ and $a-6b+6c$ are coplanar, then the direction cosines of the vector $\lambda \hat{i}-2\lambda \hat{j}+\hat{k}$ are

If the vertices of a tetrahedron are $\vec{a} = \vec{j} + 2\vec{k}$,$\vec{b} = 3\vec{i} + \vec{k}$,$\vec{c} = 4\vec{i} + 3\vec{j} + 6\vec{k}$,and $\vec{d} = 2\vec{i} + 3\vec{j} + 2\vec{k}$,find its volume.