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 $\hat{i}-3 \hat{j}+\hat{k}$ 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 $\vec{c} = (0, 1, 0)$:

Which of the following expressions is meaningful?

If $[\vec{p}-\vec{r}, \vec{q}, \vec{s}] + [\vec{p}+\vec{q}, \vec{r}, \vec{s}] = m[\vec{p}, \vec{r}, \vec{s}] + n[\vec{q}, \vec{r}, \vec{s}] + t[\vec{p}, \vec{q}, \vec{s}]$,then the values of $m$,$n$,$t$ respectively are . . . . . .

If $\vec{OA}=6 \hat{i}+3 \hat{j}-4 \hat{k}$,$\vec{OB}=2 \hat{j}+\hat{k}$,and $\vec{OC}=5 \hat{i}-\hat{j}+2 \hat{k}$ are the coterminous edges of a parallelepiped,then the height of the parallelepiped drawn from the vertex $A$ is

$A$ unit vector $\vec{e} = a \hat{i} + b \hat{j} + c \hat{k}$ is coplanar with the vectors $\hat{i} - 3 \hat{j} + 5 \hat{k}$ and $3 \hat{i} + \hat{j} - 5 \hat{k}$. If $\vec{e}$ is perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$,then $2 a^2 + 3 b^2 + 4 c^2 =$