If four distinct points with position vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are coplanar,then $[\vec{a} \vec{b} \vec{c}]$ is equal to

If $a = i + j + k$,$b = 4i + 3j + 4k$ and $c = i + \alpha j + \beta k$ are linearly dependent vectors and $|c| = \sqrt{3}$,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 $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 3\hat{k}$ and $\vec{c}$ is a unit vector, then the maximum value of the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is

If the points whose position vectors are $3i - 2j - k,$ $2i + 3j - 4k,$ $-i + j + 2k,$ and $4i + 5j + \lambda k$ lie on a plane,then $\lambda = $

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 . . . . . .