Let the volume of the tetrahedron with vertices $\hat{i}-\hat{j}-2\hat{k}$, $-2\hat{i}+\hat{j}-2\hat{k}$, $-\hat{i}-2\hat{j}+\hat{k}$, and $2\hat{i}+2\hat{j}+a\hat{k}$ be $\frac{20}{3}$. Then the integral value of $a$ is

The volume of the parallelepiped whose conterminous edges are $\vec{a} = \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} - 4\hat{j} + 5\hat{k}$,and $\vec{c} = 3\hat{i} - 5\hat{j} + 2\hat{k}$ is:

The position vectors of the points $A, B, C$ and $D$ are $3 \hat{i}-2 \hat{j}-\hat{k}, 2 \hat{i}-3 \hat{j}+2 \hat{k}, \hat{i}-\hat{j}+2 \hat{k}$ and $4 \hat{i}-\hat{j}-\lambda \hat{k}$ respectively. If the points $A, B, C$ and $D$ lie on a plane, the value of $\lambda$ is

Let $a = i - k$, $b = xi + j + (1 - x)k$, and $c = yi + xj + (1 + x - y)k$. Then $[a\,b\,c]$ depends on

If the points having the position vectors $-\hat{i}+4 \hat{j}-4 \hat{k}$,$3 \hat{i}+2 \hat{j}-5 \hat{k}$,$-3 \hat{i}+8 \hat{j}-5 \hat{k}$ and $-3 \hat{i}+2 \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda=$

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})] = $