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

If the vectors $\overrightarrow{a}=\hat{i}+a \hat{j}+a^{2} \hat{k}$,$\overrightarrow{b}=\hat{i}+b \hat{j}+b^{2} \hat{k}$ and $\overrightarrow{c}=\hat{i}+c \hat{j}+c^{2} \hat{k}$ are three non-coplanar vectors and $\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0$,then the value of $abc$ is

If the vectors $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$,$\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}$,and $\vec{c}=3 \hat{i}+p \hat{j}+5 \hat{k}$ are coplanar,then $p=$

The volume of the tetrahedron having vertices $(1, -6, 10)$,$(-1, -3, 7)$,$(5, -1, \lambda)$ and $(7, -4, 7)$ is $11 \text{ cubic units}$. Then $\lambda = $

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 $\bar{V} = 2\bar{i} + \bar{j} - \bar{k}$ 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 ...