If the vectors $4i+11j+mk$,$7i+2j+6k$,and $i+5j+4k$ are coplanar,then $m$ is

The volume of a tetrahedron with coterminus edges $\bar{a}, \bar{b}, \bar{c}$ is $\frac{64}{3}$ cubic units. Then,the volume of a parallelepiped with coterminus edges given by the vectors $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$ is ... cubic units.

If $\overline{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\overline{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$,then the value of $(2 \bar{a}-\bar{b}) \cdot [(\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b})] = $

$a$ is perpendicular to both $b$ and $c$. The angle between $b$ and $c$ is $\frac{2 \pi}{3}$. If $|a|=2$, $|b|=3$, and $|c|=4$, then $c \cdot (a \times b)$ is equal to (in $\sqrt{3}$)

If the volume of a tetrahedron whose vertices are $A \equiv (1, -6, 10)$,$B \equiv (-1, -3, 7)$,$C \equiv (5, -1, k)$,and $D \equiv (7, -4, 7)$ is $11$ cubic units,then the value of $k$ is:

If $\overline{a}=\hat{i}-\hat{k}$,$\overline{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\overline{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$,then $\overline{a} \cdot(\overline{b} \times \overline{c})$ depends on