If $a, b$ and $c$ are non-coplanar vectors and the four points with position vectors $2a+3b-c$,$a-2b+3c$,$3a+4b-2c$ and $ka-6b+6c$ are coplanar,then $k=$

If $d = \lambda (a \times b) + \mu (b \times c) + \nu (c \times a)$ and $[a, b, c] = \frac{1}{8}$,then $\lambda + \mu + \nu$ is equal to

Let $a, b$ and $c$ be distinct non-negative numbers. If the vectors $a\hat{i} + a\hat{j} + c\hat{k}$,$\hat{i} + \hat{k}$ and $c\hat{i} + c\hat{j} + b\hat{k}$ are coplanar,then $c = \dots$

$a \cdot (b \times c)$ 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:

The volume of a parallelepiped whose coterminous edges are represented by the vectors $\overrightarrow{OA} = (2, 1, 1)$,$\overrightarrow{OB} = (3, -1, 1)$,and $\overrightarrow{OC} = (-1, 1, -1)$ is . . . . . . cubic units.