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:

The vector $c \cdot (b+c) \times (a+b+c)$ is equal to

Let $\bar{a}=\lambda \bar{i}+3 \bar{j}+4 \bar{k}$, $\bar{b}=3 \bar{i}-\bar{j}+\lambda \bar{k}$ and $\bar{c}=\lambda \bar{i}+\bar{j}-3 \bar{k}$ be three vectors for some integer $\lambda$. If the volume of the parallelepiped with $\bar{a}, \bar{b}, \bar{c}$ as coterminus edges is $61$ cubic units, then the number of possible values of $\lambda$ is

Let $\overrightarrow{a}=\hat{i}-2 \hat{j}$,$\overrightarrow{b}=2 \hat{j}+3 \hat{k}$,$\overrightarrow{c}=p\hat{i}+q \hat{j}$ and $\overrightarrow{d}=p \hat{j}-q \hat{k}$ be four vectors. If $(\vec{a} \times \vec{b}) \cdot \vec{c}=3=(\vec{a} \times \vec{b}) \cdot \vec{d}$,then $3 p+q=$

If $\vec{u} = \hat{i} + 2\hat{j} - 2\hat{k}$, $\vec{v} = 2\hat{i} + \hat{k}$ and $\vec{w}$ is a unit vector, then the maximum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is

If $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$,and $\vec{c} = \lambda\hat{i} + \hat{j} + (2\lambda - 1)\hat{k}$ are coplanar vectors,then $\lambda$ is equal to: