If three conterminous edges of a parallelepiped are represented by $\vec{a} - \vec{b}$,$\vec{b} - \vec{c}$,and $\vec{c} - \vec{a}$,then its volume is

If $\overline{a}, \overline{b}, \overline{c}$ are mutually perpendicular vectors having magnitudes $1, 2, 3$ respectively,then the value of $[\overline{a}+\overline{b}+\overline{c} \quad \overline{b}-\overline{a} \quad \overline{c}]$ is

If $a, b, c$ are non-coplanar vectors and $d = \lambda a + \mu b + \nu c$,then $\lambda = \dots$

If the vectors $a\hat{i}+\hat{j}+\hat{k}$,$\hat{i}+b\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+c\hat{k}$ are coplanar $(a \neq 1, b \neq 1, c \neq 1)$,then the value of $abc-(a+b+c)$ is:

The volume of the parallelepiped whose coterminous edges are $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = 3\hat{i} - \hat{j} - \hat{k}$, and $\vec{c} = \hat{j} + 3\hat{k}$ 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 $\vec{c} = (0, 1, 0)$: