If $a$ and $b$ are parallel vectors,then $[a \ c \ b] = $

If $\bar{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \bar{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$,and $\bar{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$,and $[3 \bar{a}+\bar{b} \quad 3 \bar{b}+\bar{c} \quad 3 \bar{c}+\bar{a}] = \lambda \begin{vmatrix} \bar{a} \cdot \hat{i} & \bar{a} \cdot \hat{j} & \bar{a} \cdot \hat{k} \\ \bar{b} \cdot \hat{i} & \bar{b} \cdot \hat{j} & \bar{b} \cdot \hat{k} \\ \bar{c} \cdot \hat{i} & \bar{c} \cdot \hat{j} & \bar{c} \cdot \hat{k} \end{vmatrix}$,then the value of $\lambda$ is:

If $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors and $\vec{p}, \vec{q}, \vec{r}$ are defined as $\vec{p} = \frac{\vec{b} \times \vec{c}}{[\vec{a} \vec{b} \vec{c}]}, \vec{q} = \frac{\vec{c} \times \vec{a}}{[\vec{a} \vec{b} \vec{c}]}, \vec{r} = \frac{\vec{a} \times \vec{b}}{[\vec{a} \vec{b} \vec{c}]}$, then $[(\vec{a} + \vec{b}) \cdot \vec{p} + (\vec{b} + \vec{c}) \cdot \vec{q} + (\vec{c} + \vec{a}) \cdot \vec{r}]$ is equal to

For any non-zero vectors $\bar{a}$ and $\bar{b}$,$\left[\begin{array}{lll}\bar{b} & \bar{a} \times \bar{b} & \bar{a}\end{array}\right]=$

If $\overrightarrow{a} = \alpha \hat{i} + \beta \hat{j} + 3 \hat{k}$,$\overrightarrow{b} = -\beta \hat{i} - \alpha \hat{j} - \hat{k}$,and $\overrightarrow{c} = \hat{i} - 2 \hat{j} - \hat{k}$ such that $\overrightarrow{a} \cdot \overrightarrow{b} = 1$ and $\overrightarrow{b} \cdot \overrightarrow{c} = -3$,then $\frac{1}{3}((\vec{a} \times \vec{b}) \cdot \vec{c})$ is equal to ............

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