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 $$\mathrm{p}=$$

If the line $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$$ meets the plane $$x+2 y+3 z=15$$ at the point $$P$$, then the distance of $$\mathrm{P}$$ from the origin is

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