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})] = $

If the vectors $2 \hat{i}+3 \hat{j}+4 \hat{k}$,$2 \hat{i}+\hat{j}-\hat{k}$ and $\lambda \hat{i}-\hat{j}+2 \hat{k}$ are coplanar,then the value of $\lambda$ is:

If the vectors $a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k}, \hat{i}+\hat{j}+c \hat{k}$ $(a \neq 1, b \neq 1, c \neq 1)$ are coplanar,then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is

If $\bar{a}+\bar{b}, \bar{b}+\bar{c}$ and $\bar{c}+\bar{a}$ are coterminous edges of a parallelepiped,then its volume is $ . . . . . . $

The volume of the tetrahedron having vertices $(1, -6, 10)$,$(-1, -3, 7)$,$(5, -1, \lambda)$ and $(7, -4, 7)$ is $11 \text{ cubic units}$. Then $\lambda = $

Let $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}-\hat{k}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c}=11$,$\vec{b} \cdot(\vec{a} \times \vec{c})=27$ and $\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|$,then $|\vec{a} \times \vec{c}|^2$ is equal to