The sum of all values of $\alpha$,for which the points whose position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$,$2 \hat{i}-3 \hat{j}+4 \hat{k}$,$(\alpha+1) \hat{i}+2 \hat{k}$ and $9 \hat{i}+(\alpha-8) \hat{j}+6 \hat{k}$ are coplanar,is equal to

If $\overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are non-zero coplanar vectors,then $[2 \overrightarrow{a}-\overrightarrow{b} \quad 3 \overrightarrow{b}-\overrightarrow{c} \quad 4 \overrightarrow{c}-\overrightarrow{a}]$ is

$[b \times c, c \times a, a \times b]$ is equal to

The value of $a$,so that the volume of the parallelepiped formed by $\hat{i} + a \hat{j} + \hat{k}$,$\hat{j} + a \hat{k}$,and $a \hat{i} + \hat{k}$ becomes minimum is

If the vectors $m \hat{i} + m \hat{j} + n \hat{k}$,$\hat{i} + \hat{k}$,and $n \hat{i} + n \hat{j} + p \hat{k}$ lie in a plane,then...

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: