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 :

Three vectors $\vec{a} = \hat{i} + \hat{j}$,$\vec{b} = \hat{j} + \hat{k}$,and $\vec{c} = \hat{k} + \hat{i}$ are given. If three unit vectors are drawn perpendicular to the three planes formed by these vectors,what is the volume of the parallelepiped formed by these unit vectors?

If the vectors $(1 - x)\hat i + \hat j + \hat k$,$\hat i + (1 - y)\hat j + \hat k$,and $\hat i + \hat j + (1 - z)\hat k$ are coplanar,then the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is $(x, y, z \neq 0)$.

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:

Let $\alpha \in \mathbb{R}$ and the three vectors $\vec{a} = \alpha \hat{i} + \hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + \hat{j} - \alpha \hat{k}$,and $\vec{c} = \alpha \hat{i} - 2\hat{j} + 3\hat{k}$. Then the set $S = \{ \alpha : \vec{a}, \vec{b}, \text{ and } \vec{c} \text{ are coplanar} \}$

If $\bar{u}=\hat{\imath}-2 \hat{\jmath}+\hat{k}, \bar{v}=3 \hat{\imath}+\hat{k}$ and $\bar{w}=\hat{\jmath}-\hat{k}$,then the volume of the parallelepiped with $\bar{u} \times \bar{v}, \bar{u}+\bar{w}$ and $\bar{v}+\bar{w}$ as coterminus edges is