If $a, b,$ and $c$ are coplanar unit vectors,find the value of the scalar triple product $[2a - b, 2b - c, 2c - a]$.

$(\overrightarrow{a}+2 \overrightarrow{b}-\overrightarrow{c}) \cdot ((\overrightarrow{a}-\overrightarrow{b}) \times (\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{c}))$ is equal to

The volume of a parallelepiped,whose coterminous edges are given by $\bar{u}=\hat{i}+\hat{j}+\lambda \hat{k}$,$\bar{v}=\hat{i}+\hat{j}+3 \hat{k}$,and $\bar{w}=2 \hat{i}+\hat{j}+\hat{k}$,is $1$ cubic unit. If $\theta$ is the angle between $\bar{u}$ and $\bar{w}$,then the value of $\cos \theta$ is:

If the vectors $2\hat{i} - \hat{j} + \hat{k}$,$\hat{i} + 2\hat{j} - 3\hat{k}$,and $3\hat{i} + a\hat{j} + 5\hat{k}$ are coplanar,then the value of $a$ is:

The vectors $i + 2j + 3k$,$\lambda i + 4j + 7k$,and $-3i - 2j - 5k$ are collinear if $\lambda$ equals:

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector,assuming the question implies the vectors are linearly dependent or part of a coplanar set,find $\lambda$. Given the standard form of such problems,if we consider the vectors $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ to be coplanar with a reference vector,let us assume the third vector is $\hat{k}$. For these to be coplanar,the scalar triple product must be zero: $\left|\begin{array}{ccc} 1 & -3 & 1 \\ \lambda & 3 & 0 \\ 0 & 0 & 1 \end{array}\right| = 0$. Solving this,$\lambda$ is equal to: