Three vectors $\vec{a} = \hat{i} + \hat{j}$,$\vec{b} = \hat{j} + \hat{k}$,and $\vec{c} = \hat{k} + \hat{i}$ are given. If three unit vectors are drawn perpendicular to the three planes formed by these vectors,what is the volume of the parallelepiped formed by these unit vectors?

If the vector $\overline{c}$ lies in the plane of $\overline{a}$ and $\overline{b}$,where $\overline{a}=\hat{i}-\hat{j}+2\hat{k}$,$\overline{b}=\hat{i}+\hat{j}+\hat{k}$ and $\overline{c}=x\hat{i}-(2-x)\hat{j}-\hat{k}$,then the value of $x$ is

The value of $[a - b, b - c, c - a]$,where $|a| = 1, |b| = 5$ and $|c| = 3$,is

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector, let us assume the vectors are $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$, $\vec{b} = \lambda \hat{i}+3 \hat{j}$, and we consider the standard basis vectors or a third vector to define coplanarity. However, if the question implies these two vectors are coplanar with the origin or a specific plane, we evaluate the scalar triple product. Given the standard interpretation of such problems, if $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ are coplanar with $\vec{c} = \hat{j}$, then the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 0$. Solving for $\lambda$ where $\vec{a} = (1, -3, 1)$, $\vec{b} = (\lambda, 3, 0)$, and $\vec{c} = (0, 1, 0)$:

$a$ is perpendicular to both $b$ and $c$. The angle between $b$ and $c$ is $\frac{2 \pi}{3}$. If $|a|=2$, $|b|=3$, and $|c|=4$, then $c \cdot (a \times b)$ is equal to (in $\sqrt{3}$)

$a, b, c$ 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