The equation of the plane through $$(-1,1,2)$$ whose normal makes equal acute angles with co-ordinate axes is

The distance of the point $$\mathrm{P}(-2,4,-5)$$ from the line $$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$$ is

If the line $$\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$$ and $$\frac{7-7 x}{3 \mathrm{p}}=\frac{y-5}{1}=\frac{6-\mathrm{z}}{5}$$ are at right angles, then $$\mathrm{p}=$$