Consider the set of eight vectors $V=\{a \hat{i}+b \hat{j}+c \hat{k}: a, b, c \in\{-1,1\}\}$. Three non-coplanar vectors can be chosen from $V$ in $2^p$ ways. Then $p$ is

If the given vectors $(-bc, b^2 + bc, c^2 + bc)$,$(a^2 + ac, -ac, c^2 + ac)$ and $(a^2 + ab, b^2 + ab, -ab)$ are coplanar,where none of $a, b$ and $c$ is zero,then:

For what value of $\lambda$ is the volume of the tetrahedron with vertices having position vectors $\hat{i} - 6\hat{j} + 10\hat{k}$,$-\hat{i} - 3\hat{j} + 7\hat{k}$,$5\hat{i} - \hat{j} + \lambda\hat{k}$,and $7\hat{i} - 4\hat{j} + 7\hat{k}$ equal to $11$ cubic units?

Let $\vec{v}=\alpha \hat{i}+2 \hat{j}-3 \hat{k}$,$\vec{w}=2 \alpha \hat{i}+\hat{j}-\hat{k}$,and $\vec{u}$ be a vector such that $|\vec{u}|=\alpha > 0$. If the minimum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is $-\alpha \sqrt{3401}$,and $|\vec{u} \cdot \hat{i}|^2=\frac{m}{n}$ where $m$ and $n$ are coprime natural numbers,then $m + n$ is equal to $.........$.

If $\overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are non-zero coplanar vectors,then $[2 \overrightarrow{a}-\overrightarrow{b} \quad 3 \overrightarrow{b}-\overrightarrow{c} \quad 4 \overrightarrow{c}-\overrightarrow{a}]$ is

If $[a, b, c] = 3$, then the volume (in cubic units) of the parallelepiped with $2a+b$, $2b+c$, and $2c+a$ as edges is: