Let $A=(2,0,-1)$,$B=(1,-2,0)$,$C=(1,2,-1)$,and $D=(0,-1,-2)$ be four points. If $\theta$ is the acute angle between the plane determined by $A, B, C$ and the plane determined by $A, C, D$,then $\tan \theta=$

Show that the points $(\hat{i}-\hat{j}+3 \hat{k})$ and $3(\hat{i}+\hat{j}+\hat{k})$ are equidistant from the plane $\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$ and lie on opposite sides of it.

Let $A$ be a point having position vector $\bar{i}-3 \bar{j}$ and $\bar{r}=(\bar{i}-3 \bar{j})+t(\bar{j}-2 \bar{k})$ be a line. If $P$ is a point on this line and is at a minimum distance from the plane $\bar{r} \cdot(2 \bar{i}+3 \bar{j}+5 \bar{k})=0$, then the equation of the plane through $P$ and perpendicular to $AP$ is:

The distance between the line $r = 2i - 2j + 3k + \lambda (i - j + 4k)$ and the plane $r \cdot (i + 5j + k) = 5$ is

The sine of the angle between the straight line $\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}$ and the plane $2x-2y+z=5$ is

Let a plane $P$ contain two lines $\overrightarrow{r} = \hat{i} + \lambda(\hat{i} + \hat{j}), \lambda \in R$ and $\overrightarrow{r} = -\hat{j} + \mu(\hat{j} - \hat{k}), \mu \in R$. If $Q(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from the point $M(1, 0, 1)$ to $P$,then $3(\alpha + \beta + \gamma)$ equals