Let L be the line of intersection of the planes $2 x+3 y+z=1$ and $x+3 y+2 z=2$. If L makes an angle $\alpha$ with the positive X -axis, then $\cos \alpha$ equals

The equation of the plane, passing through the point $(1,1,1)$ and perpendicular to the planes $2 x+y-2 z=5$ and $3 x-6 y-2 z=7$, is

The distance of the point $(1,3,-7)$ from the plane passing through the point $(1,-1,-1)$ having normal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$ is

The value of m , such that $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-m}{2}$ lies in the plane $2 x-4 y+z=7$, is