If the volume of the parallelepiped is $158 \text{ cubic units}$,whose coterminous edges are given by the vectors $\bar{a} = (\hat{i} + \hat{j} + n \hat{k})$,$\bar{b} = (2 \hat{i} + 4 \hat{j} - n \hat{k})$,and $\bar{c} = (\hat{i} + n \hat{j} + 3 \hat{k})$,where $n \geq 0$,then the value of $n$ is:

If $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$,and $\vec{c} = \lambda\hat{i} + \hat{j} + (2\lambda - 1)\hat{k}$ are coplanar vectors,then $\lambda = . . . .$

Let the volume of a parallelepiped whose coterminous edges are given by $\overrightarrow{u}=\hat{i}+\hat{j}+\lambda \hat{k}$,$\overrightarrow{v}=\hat{i}+\hat{j}+3 \hat{k}$ and $\overrightarrow{w}=2 \hat{i}+\hat{j}+\hat{k}$ be $1 \text{ cu. unit}$. If $\theta$ is the angle between the edges $\overrightarrow{u}$ and $\overrightarrow{w}$,then $\cos \theta$ can be

The value of $[a - b, b - c, c - a]$,where $|a| = 1, |b| = 5$ and $|c| = 3$,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:

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points $A, B, C$ and $D$ be $\vec{a}-\vec{b}+\vec{c}$,$\lambda \vec{a}-3 \vec{b}+4 \vec{c}$,$-\vec{a}+2 \vec{b}-3 \vec{c}$ and $2 \vec{a}-4 \vec{b}+6 \vec{c}$ respectively. If $\overrightarrow{AB}$,$\overrightarrow{AC}$ and $\overrightarrow{AD}$ are coplanar,then $\lambda$ is :