If $a, b, c$ are non-coplanar vectors and $\lambda$ is a real number,then $[\lambda(a + b), \lambda^2 b, \lambda c] = [a, b + c, b]$ for

The maximum volume of a parallelepiped (in cubic units) with vectors $\vec{u} = 2a\hat{i} + \hat{k}$, $\vec{v} = a\hat{j} - a\hat{k}$, and $\vec{w} = 3\hat{i} + a\hat{j}$, where $a \in [0, 1]$, as its coterminous edges is:

The volume of the parallelepiped determined by the vectors $\vec{a} + \vec{b}, \vec{b} + \vec{c}$ and $\vec{c} + \vec{a}$ is $4$. Then the volume of the parallelepiped determined by the vectors $\vec{a} \times \vec{b}, \vec{b} \times \vec{c}$ and $\vec{c} \times \vec{a}$ 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 $\vec{c} = (0, 1, 0)$:

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

If the volume of a tetrahedron whose conterminous edges are $\overline{a}+\overline{b}, \overline{b}+\overline{c}, \overline{c}+\overline{a}$ is $24$ cubic units,then the volume of the parallelepiped whose coterminous edges are $\overline{a}, \overline{b}, \overline{c}$ is