If $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$,then $[\vec{a} \vec{b} \vec{c}]$ depends on

If the volume of the parallelepiped formed by three non-coplanar vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $4$ cubic units,then $[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}]$ is equal to

Let $a, b, c$ be distinct non-negative numbers. If the vectors $a\hat{i} + a\hat{j} + c\hat{k}$,$\hat{i} + \hat{k}$,and $c\hat{i} + c\hat{j} + b\hat{k}$ lie in a plane,then $c$ is

Let $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{B} = \hat{i}$,and $\overrightarrow{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$. If $C_2 = -1$ and $C_3 = 1$,then to make the three vectors coplanar:

If $\vec{a} = \hat{i} - \hat{k}$, $\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}$ and $\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$, then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ depends on

Let the vectors $\vec{u} = (2+a+b) \hat{i}+(a+2 b+c) \hat{j}-(b+c) \hat{k}$,$\vec{v} = (1+b) \hat{i}+2 b \hat{j}-b \hat{k}$,and $\vec{w} = (2+b) \hat{i}+2 b \hat{j}+(1-b) \hat{k}$ where $a, b, c \in \mathbb{R}$ be co-planar. Then which of the following is true?