If $\vec{u} = \hat{j} + 4\hat{k}$,$\vec{v} = \hat{i} + 3\hat{k}$ and $\vec{w} = \cos \theta \hat{i} + \sin \theta \hat{j}$ are vectors in $3$-dimensional space,then the maximum possible value of $|(\vec{u} \times \vec{v}) \cdot \vec{w}|$ is

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:

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:

$|(a \times b) \cdot c| = |a| |b| |c|$,if

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

The volume (in cubic units) of the tetrahedron with edges $\hat{i}+\hat{j}+\hat{k}$, $\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}-\hat{k}$ is