If the vectors $(1 - x)\hat i + \hat j + \hat k$,$\hat i + (1 - y)\hat j + \hat k$,and $\hat i + \hat j + (1 - z)\hat k$ are coplanar,then the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is $(x, y, z \neq 0)$.

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

Let $\vec{v} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{w} = \hat{i} + 3\hat{k}$. If $\vec{u}$ is a unit vector,then the maximum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is

For any non-zero vectors $a, b, c$,$a \cdot[(b+c) \times(a+b+c)] = \ldots .$

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 $a, b, c$ are non-coplanar vectors and $d = \lambda a + \mu b + \nu c$,then $\lambda$ is equal to