For what value of $\lambda$ are the vectors $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = \lambda\hat{i} + 4\hat{j} + 7\hat{k}$,and $\vec{c} = -3\hat{i} - 2\hat{j} - 5\hat{k}$ coplanar?

If $(2,3,9), (5,2,1), (1, \lambda, 8)$ and $(\lambda, 2,3)$ are coplanar,then the product of all possible values of $\lambda$ is.

If $\bar{a}, \bar{b},$ and $\bar{c}$ are mutually perpendicular vectors such that $|\bar{a}| = 1, |\bar{b}| = 3,$ and $|\bar{c}| = 5,$ then find the value of $[\bar{a} - 2\bar{b}, \bar{b} - 3\bar{c}, \bar{c} - 4\bar{a}]$.

Let $S$ be the set of all $(\lambda, \mu)$ for which the vectors $\lambda \hat{i} - \hat{j} + \hat{k}$,$\hat{i} + 2\hat{j} + \mu \hat{k}$ and $3\hat{i} - 4\hat{j} + 5\hat{k}$,where $\lambda - \mu = 5$,are coplanar,then $\sum_{(\lambda, \mu) \in S} 80(\lambda^2 + \mu^2)$ 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 $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$ be vectors in three-dimensional space,where $\overrightarrow{u}$ and $\overrightarrow{v}$ are unit vectors which are not perpendicular to each other and $\overrightarrow{u} \cdot \overrightarrow{w}=1, \overrightarrow{v} \cdot \overrightarrow{w}=1, \overrightarrow{w} \cdot \overrightarrow{w}=4$. If the volume of the parallelepiped,whose adjacent sides are represented by the vectors $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$,is $\sqrt{2}$,then the value of $|3\vec{u}+5\vec{v}|$ is.