$A(2,3,4), B(4,5,7), C(2,-6,3), D(4,-4, k)$ are four points. If the line $\overline{AB}$ is parallel to $\overline{CD}$,then $k$ is equal to

Let the vertices $Q$ and $R$ of the triangle $PQR$ lie on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$. Given $QR=5$ and the coordinates of the point $P$ are $(0,2,3)$. If the area of the triangle $PQR$ is $\frac{m}{n}$,then:

The lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are perpendicular to each other. Then,the value of $p$ is . . . . . . .

If the shortest distance between the lines $\frac{x-k}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$,then $k=$

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

The angle between a line with direction ratios $2, 2, 1$ and the line joining the points $(3, 1, 4)$ and $(7, 2, 12)$ is