If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3 y-\alpha z+\beta=0$, then $(\alpha, \beta)=$

A plane which is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, passes through $(1,2,1)$. The distance of the plane from the point $(2,3,4)$ 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

A line with positive direction cosines passes through the point $\mathrm{P}(2,1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2 x+y+\mathrm{z}=9$ at point Q . The length of the line segment PQ equals $\qquad$ units.