The edges of a parallelepiped are of unit length and are parallel to non-coplanar unit vectors $\hat{a}, \hat{b}, \hat{c}$ such that $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = 1/2$. Then the volume of the parallelepiped is

The number of integral values of $p$ for which the vectors $(p+1) \hat{i} - 3 \hat{j} + p \hat{k}$,$p \hat{i} + (p+1) \hat{j} - 3 \hat{k}$,and $-3 \hat{i} + p \hat{j} + (p+1) \hat{k}$ are linearly dependent is:

Consider the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-6 \hat{k}$,$\vec{b}=6 \hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{c}=3 \hat{i}-6 \hat{j}-2 \hat{k}$.
Assertion $(A):$ The three vectors do not form a triangle.
Reason $(R):$ The three vectors are non-coplanar.
The correct option among the following is:

Let $\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$. Let $\vec{d}$ be a vector which is perpendicular to both $\vec{a}$ and $\vec{b}$,and $\vec{c} \cdot \vec{d}=12$. Then $(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$ is equal to $........$.

Let $v = 2i + j - k$ and $w = i + 3k$. If $u$ is any unit vector,then the maximum value of the scalar triple product $[u v w]$ is

Let $\vec{u} = a\hat{i} + b\hat{j} + c\hat{k}$,$\vec{v} = b\hat{i} + c\hat{j} + a\hat{k}$,and $\vec{w} = c\hat{i} + a\hat{j} + b\hat{k}$. If $[\vec{u} \, \vec{v} \, \vec{w}] = 0$ and $\vec{w} = \lambda \vec{x} + \mu \vec{y}$ where $(a + b + c) \neq 0$ and $\lambda, \mu \neq 0$,then the vectors $\vec{x}, \vec{y}, \vec{u}, \vec{v}, \vec{w}$ are: