The value of $\alpha$,so that the volume of the parallelepiped formed by $\hat{i}+\alpha \hat{j}+\hat{k}$,$\hat{j}+\alpha \hat{k}$,and $\alpha \hat{i}+\hat{k}$ becomes maximum,is

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 three conterminous edges of a parallelepiped are represented by $\vec{a} - \vec{b}$,$\vec{b} - \vec{c}$,and $\vec{c} - \vec{a}$,then its volume is

For any non-zero vectors $a, b, c$,$a \cdot[(b+c) \times(a+b+c)] = \ldots .$

If $a, b, c$ are three coplanar vectors,then $[a + b, b + c, c + a] = $

Given the vectors $\vec x = 3i - 6j - k$,$\vec y = i + 4j - 3k$,and $\vec z = 3i - 4j - 12k$,find the projection of the vector $\vec x \times \vec y$ onto the vector $\vec z$.