If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors,then $\frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{\vec{c} \cdot (\vec{a} \times \vec{b})} + \frac{\vec{b} \cdot (\vec{a} \times \vec{c})}{\vec{c} \cdot (\vec{a} \times \vec{b})} = \dots$

The value of $a$ such that the volume of the parallelepiped formed by the vectors $i + aj + k$,$j + ak$,and $ai + k$ is minimum is:

The value of $\lambda$ for which the four points $2i + 3j - k$,$i + 2j + 3k$,$3i + 4j - 2k$,and $i - \lambda j + 6k$ are coplanar is:

Let $\alpha \in \mathbb{R}$ and the three vectors $\vec{a} = \alpha \hat{i} + \hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + \hat{j} - \alpha \hat{k}$,and $\vec{c} = \alpha \hat{i} - 2\hat{j} + 3\hat{k}$. Then the set $S = \{ \alpha : \vec{a}, \vec{b}, \text{ and } \vec{c} \text{ are coplanar} \}$

If $a, b, c, d$ are $4$ vectors such that $a \cdot b = 0$, $|a \times c| = |a||c|$, and $|a \times d| = |a||d|$, then $[b c d] = $

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 $\vec{c} = (0, 1, 0)$: