If the volume of a tetrahedron,whose vertices are $A(1, 2, 3)$,$B(-3, -1, 1)$,$C(2, 1, 3)$,and $D(-1, 2, x)$ is $\frac{11}{6}$ cubic units,then the value of $x$ is:

Let $V = 2\hat{i} + \hat{j} - \hat{k}$ and $W = \hat{i} + 3\hat{k}$. If $U$ is a unit vector,then the maximum value of $[U V W]$ 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 $\vec{c} = (0, 1, 0)$:

The lines $\overline{r}=\overline{a}+\lambda(\overline{b} \times \overline{c})$ and $\overline{r}=\overline{c}+\mu(\overline{a} \times \overline{b})$ will intersect if

If the vectors $\overrightarrow{p}=(a+1) \hat{i}+a \hat{j}+a \hat{k}$,$\overrightarrow{q}=a \hat{i}+(a+1) \hat{j}+a \hat{k}$,and $\overrightarrow{r}=a \hat{i}+a \hat{j}+(a+1) \hat{k}$ $(a \in R)$ are coplanar and $3(\overrightarrow{p} \cdot \overrightarrow{q})^{2}-\lambda|\overrightarrow{r} \times \overrightarrow{q}|^{2}=0$,then the value of $\lambda$ is:

If $a = 2i + j - k$,$b = i + 2j + k$,and $c = i - j + 2k$,then $a \cdot (b \times c) = \dots$