What will be the volume of the parallelepiped whose coterminous edges are given by the vectors $a = i - j + k$,$b = i - 3j + 4k$,and $c = 2i - 5j + 3k$?

If the points whose position vectors are $3i - 2j - k,$ $2i + 3j - 4k,$ $-i + j + 2k,$ and $4i + 5j + \lambda k$ lie on a plane,then $\lambda = $

Let $\vec{c}$ be a vector coplanar with the unit vectors $\vec{a}$ and $\vec{b}$, and let $\vec{d}$ be the unit vector perpendicular to $\vec{a}$, $\vec{b}$, and $\vec{c}$. If $[\vec{a} \vec{b} \vec{d}] \vec{c} - [\vec{a} \vec{b} \vec{c}] \vec{d} = \hat{i} + 2\hat{j} + 2\hat{k}$ and the angle between $\vec{a}$ and $\vec{b}$ is $30^{\circ}$, then $|\vec{c}| =$

Let $\vec{\lambda} = x\vec{a} + y\vec{b} + z\vec{c}$ and $\vec{\lambda} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = 2(x + y + z)$ (where $x + y + z \neq 0$),then the scalar triple product $[\vec{a} \, \vec{b} \, \vec{c}]$ is:

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector,assuming the question implies the vectors are linearly dependent or part of a coplanar set,find $\lambda$. Given the standard form of such problems,if we consider the vectors $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ to be coplanar with a reference vector,let us assume the third vector is $\hat{k}$. For these to be coplanar,the scalar triple product must be zero: $\left|\begin{array}{ccc} 1 & -3 & 1 \\ \lambda & 3 & 0 \\ 0 & 0 & 1 \end{array}\right| = 0$. Solving this,$\lambda$ is equal to:

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 :