The equation of the line passing through the point $$(-1,3,-2)$$ and perpendicular to each of the lines $$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$$ and $$\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$$ is

If $$A(1,4,2)$$ and $$C(5,-7,1)$$ are two vertices of triangle $$A B C$$ and $$G\left(\frac{4}{3}, 0, \frac{-2}{3}\right)$$ is centroid of the triangle $$A B C$$, then the mid point of side $$B C$$ is

The distance of the point $$(-1,-5,-10)$$ from the point of intersection of the line $$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$$ and the plane $$x-y+z=5$$ is

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