If $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$,and $\vec{c} = \lambda\hat{i} + \hat{j} + (2\lambda - 1)\hat{k}$ are coplanar vectors,then $\lambda$ is equal to:

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

If $\bar{V} = 2\bar{i} + \bar{j} - \bar{k}$ and $\bar{W} = \bar{i} + 3\bar{k}$,and if $\bar{U}$ is a unit vector,then the maximum value of $[\bar{U} \bar{V} \bar{W}]$ is ...

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 $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors and $\vec{r}$ is any vector,then $[\vec{b} \, \vec{c} \, \vec{r}] \vec{a} + [\vec{c} \, \vec{a} \, \vec{r}] \vec{b} + [\vec{a} \, \vec{b} \, \vec{r}] \vec{c} = \dots$

If $[\bar{a} \times \bar{b} \quad \bar{b} \times \bar{c} \quad \bar{c} \times \bar{a}] = \lambda [\bar{a} \quad \bar{b} \quad \bar{c}]^2$,then $\lambda$ is equal to