The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar,if

Let the line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $AB$ from the plane $2x-2y+z=14$ is

Let a unit vector $\hat{OP}$ make angles $\alpha, \beta, \gamma$ with the positive directions of the coordinate axes $OX, OY, OZ$ respectively,where $\beta \in (0, \frac{\pi}{2})$. If $\hat{OP}$ is perpendicular to the plane passing through the points $(1, 2, 3)$,$(2, 3, 4)$,and $(1, 5, 7)$,then which one of the following is true?

The position vectors of points $A$ and $B$ are $i - j + 3k$ and $3i + 3j + 3k$ respectively. The equation of a plane is $r \cdot (5i + 2j - 7k) + 9 = 0$. The points $A$ and $B$:

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 $\tan \theta=$

For non-coplanar vectors $a, b$ and $c$, if the point of intersection of the line $r=a+t(b-c)$ and the plane $r=b+c+x(a-b)+y(c+a)$ is $l a+m b+n c$, then $3 l+4 m+2 n=$