The equation of the plane passing through the point of intersection of the planes $2 x-y+z-3=0$ and $4 x-3 y+5 z+9=0$ and parallel to the line $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}$ is $\alpha x+\beta y+\gamma z+d=0$ Then $\alpha+\beta+\gamma+d=$

The distance of the point $(2,4,0)$ from the point of intersection of the lines $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ and $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ is

The co-ordinates of the point where the line joining the points $(2,-3,1)$ and $(3,-4,-5)$ and intersects the plane $2 x+y+z=7$ are

If the line $\frac{x+1}{3}=\frac{y-k}{7}=\frac{z-4}{8}$ lies in the plane $2 x+\mathrm{p} y+7 z-41=0$ which is perpendicular to the plane $x+4 y-2 z+13=0$ then $\mathrm{k}=$