The equation of plane through the point $(2,-1,-3)$ and parallel to lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is

The equation of the plane, passing through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4=0$ and parallel to $Y$-axis is

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

If the distance between the plane Ax-2y+z $=\mathrm{d}$ and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6}$ units, then $|d|$ is