Find the volume of a tetrahedron whose vertices are given by the vectors $-i + j + k$,$i - j + k$,and $i + j - k$,with the fourth vertex being the origin.

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i}$, and $\vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}$ with $c_1 = 1$ and $c_2 = 2$, then find the value of $c_3$ such that $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar.

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $(\vec{a} - \lambda \vec{b}) \cdot (\vec{b} - 2\vec{c}) \times (\vec{c} + 2\vec{a}) = 0$,then $\lambda$ is equal to

If $a = 3i - 2j + 2k$,$b = 6i + 4j - 2k$ and $c = 3i - 2j - 4k$,then $a \cdot (b \times c)$ is

$\vec{a}=2 \hat{i}-\hat{j}$, $\vec{b}=2 \hat{j}-\hat{k}$, $\vec{c}=2 \hat{k}-\hat{i}$ are three vectors and $\vec{d}$ is a unit vector perpendicular to $\vec{c}$. If $\vec{a}, \vec{b}, \vec{d}$ are coplanar vectors, then $|\vec{d} \cdot \vec{b}|=$

$a, b, c$ are three non-zero,non-coplanar vectors and $p, q, r$ are three other vectors such that $p = \frac{b \times c}{a \cdot (b \times c)}$,$q = \frac{c \times a}{a \cdot (b \times c)}$,$r = \frac{a \times b}{a \cdot (b \times c)}$. Then $[p, q, r]$ equals