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

If $2 \hat{i}-\hat{j}+3 \hat{k}$, $-12 \hat{i}-\hat{j}-3 \hat{k}$, $-\hat{i}+2 \hat{j}-4 \hat{k}$, and $\lambda \hat{i}+2 \hat{j}-\hat{k}$ are the position vectors of four coplanar points, then $\lambda=$

Let $\vec{a}=\hat{i}-\alpha \hat{j}+\beta \hat{k}$,$\vec{b}=3 \hat{i}+\beta \hat{j}-\alpha \hat{k}$ and $\vec{c}=-\alpha \hat{i}-2 \hat{j}+\hat{k}$,where $\alpha$ and $\beta$ are integers. If $\vec{a} \cdot \vec{b}=-1$ and $\vec{b} \cdot \vec{c}=10$,then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to $.....$

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 maximum volume of a parallelepiped (in cubic units) with vectors $\vec{u} = 2a\hat{i} + \hat{k}$, $\vec{v} = a\hat{j} - a\hat{k}$, and $\vec{w} = 3\hat{i} + a\hat{j}$, where $a \in [0, 1]$, as its coterminous edges is:

If $d = \lambda (a \times b) + \mu (b \times c) + \nu (c \times a)$ and $[a, b, c] = \frac{1}{8}$,then $\lambda + \mu + \nu$ is equal to