$[i, k, j] + [k, j, i] + [j, k, i]$

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

Let the position vectors of the points $A, B, C$ and $D$ be $5\hat{i}+5\hat{j}+2\lambda\hat{k}$,$\hat{i}+2\hat{j}+3\hat{k}$,$-2\hat{i}+\lambda\hat{j}+4\hat{k}$ and $-\hat{i}+5\hat{j}+6\hat{k}$. Let the set $S = \{\lambda \in \mathbb{R} : \text{The points } A, B, C \text{ and } D \text{ are coplanar}\}$. Then $\sum_{\lambda \in S}(\lambda+2)^2$ is equal to

Consider the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-6 \hat{k}$,$\vec{b}=6 \hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{c}=3 \hat{i}-6 \hat{j}-2 \hat{k}$.
Assertion $(A):$ The three vectors do not form a triangle.
Reason $(R):$ The three vectors are non-coplanar.
The correct option among the following is:

If $A, B, C$ and $D$ are $(3,7,4), (5,-2,-3), (-4,5,6)$ and $(1,2,3)$ respectively,then the volume of the parallelepiped with $AB, AC$ and $AD$ as the co-terminus edges is .... cubic units.

If the position vectors of the vertices $A, B$ and $C$ are $6i$,$6j$ and $k$ respectively with respect to the origin $O$,then the volume of the tetrahedron $OABC$ is