If the line $\frac{x+1}{3}=\frac{y-k}{7}=\frac{z-4}{8}$ lies in the plane $2 x+\mathrm{p} y+7 z-41=0$ which is perpendicular to the plane $x+4 y-2 z+13=0$ then $\mathrm{k}=$

The direction cosines of the line $x-y+2 z=5$ and $3 x+y+z=6$ are

If the angle between the planes $x-2 y+3 z-5=0$ and $x+\alpha y+2 z+7=0$ is $\cos ^{-1}\left(\frac{1}{14}\right)$ then the difference between the values of $\alpha$ is

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