The volume of the parallelepiped whose edges are represented by $-12i + \alpha k$,$3j - k$,and $2i + j - 15k$ is $546$. Then $\alpha = $

Let $\vec{u} = a\hat{i} + b\hat{j} + c\hat{k}$,$\vec{v} = b\hat{i} + c\hat{j} + a\hat{k}$,and $\vec{w} = c\hat{i} + a\hat{j} + b\hat{k}$. If $[\vec{u} \, \vec{v} \, \vec{w}] = 0$ and $\vec{w} = \lambda \vec{x} + \mu \vec{y}$ where $(a + b + c) \neq 0$ and $\lambda, \mu \neq 0$,then the vectors $\vec{x}, \vec{y}, \vec{u}, \vec{v}, \vec{w}$ are:

Let $V = 2\hat{i} + \hat{j} - \hat{k}$ and $W = \hat{i} + 3\hat{k}$. If $U$ is a unit vector,then the maximum value of $[U V W]$ is

If $a = i + j + k$,$b = 4i + 3j + 4k$,and $c = i + \alpha j + \beta k$ are coplanar vectors and $|c| = \sqrt{3}$,then:

The vectors $i + 2j + 3k$,$\lambda i + 4j + 7k$,and $-3i - 2j - 5k$ are collinear if $\lambda$ equals:

If the volume of the parallelopiped with $\vec{a} \times \vec{b}, \vec{b} \times \vec{c}$ and $\vec{c} \times \vec{a}$ as coterminous edges is $9 \text{ cu. units}$, then the volume of the parallelopiped with $(\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c}),(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})$ and $(\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})$ as coterminous edges is