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

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 $a, b, c$ are any three vectors and their reciprocal vectors are $a^{-1}, b^{-1}, c^{-1}$ such that $[a, b, c] \neq 0$,then $[a^{-1}, b^{-1}, c^{-1}]$ is equal to:

The volume of the parallelepiped whose conterminous edges are $\vec{a} = \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} - 4\hat{j} + 5\hat{k}$,and $\vec{c} = 3\hat{i} - 5\hat{j} + 2\hat{k}$ is:

Let $\overrightarrow{a}$ be a unit vector,$\overrightarrow{b} = 2\hat{i} + \hat{j} - \hat{k}$ and $\overrightarrow{c} = \hat{i} + 3\hat{k}$. Then,the maximum value of $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$ is

The volume (in cubic units) of the tetrahedron with edges $\hat{i}+\hat{j}+\hat{k}$,$\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}-\hat{k}$ is