Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\vec{c} = x\hat{i} + (x-2)\hat{j} - \hat{k}$. If the vector $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$,then $x$ equals:

The volume of the parallelepiped whose coterminous edges are represented by the vectors $2i - 3j + 4k$,$i + 2j - 2k$,and $3i - j + k$ is ............ $cubic \ units$.

If the vectors $m \hat{i} + m \hat{j} + n \hat{k}$,$\hat{i} + \hat{k}$,and $n \hat{i} + n \hat{j} + p \hat{k}$ lie in a plane,then...

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 $a(\alpha \times \beta)+b(\beta \times \gamma)+c(\gamma \times \alpha)=0$ and at least one of the scalars $a, b, c$ is non-zero,then the vectors $\alpha, \beta, \gamma$ are

If $\alpha = 2i + 3j - k$,$\beta = -i + 2j - 4k$,and $\gamma = i + j + k$,then $(\alpha \times \beta) \cdot (\alpha \times \gamma)$ is equal to