Unit vectors $a, b$ and $c$ are coplanar. $A$ unit vector $d$ is perpendicular to them. If $(a \times b) \times (c \times d) = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$ and the angle between $a$ and $b$ is $30^\circ$,then $c$ is:

Let $\vec{a} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k})$, $\vec{b} = (b_1\hat{i} + b_2\hat{j} + b_3\hat{k})$, and $\vec{c} = (c_1\hat{i} + c_2\hat{j} + c_3\hat{k})$ be three non-zero vectors such that $\vec{a}$ is a unit vector perpendicular to both $\vec{b}$ and $\vec{c}$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$, then find the value of $\left| \begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix} \right|^2$.

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...

If $x, y$ and $z$ are non-zero real numbers and $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ are such that $\vec{a} \times \vec{b}=z \hat{i}-3 \hat{j}+xy \hat{k}$ is not given,but $\vec{a} \times \vec{b}=6 \hat{i}-3 \hat{j}+\hat{k}$ is given as $z \hat{i}-3 \hat{j}+\hat{k}$,then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ 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:

If $\vec{a} = 4\hat{i} - 2\hat{j} + \hat{k}$,$\vec{b} = 3\hat{i} + 2\hat{j} - \hat{k}$,and $\vec{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ represent the three coterminous edges of a parallelepiped,find its volume.