The equation of the plane containing the line $\frac{x+1}{2}=\frac{y+2}{1}=\frac{z-2}{3}$ and the point $(1,-1,3)$ is

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 equation of a plane passing through the intersection of two planes $x+2y-3z+2=0$ and $6x+y+z+1=0$ and parallel to the line $x-1=y+2=7-z$ is

If $P=(2,-3,4)$,$Q=(-1,-4,0)$,and $R=(2,1,0)$ are three points,and $S$ is the foot of the perpendicular drawn from $R$ to the line $PQ$,then the $X$-coordinate of $S$ is:

$A$ vector parallel to the line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is

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 $\tan \theta=$