Sheaf of Planes
A sheaf of planes with an axis, r, is a set of planes that contain the line r

A_{1}x+B_{1}y+C_{1}z+D_{1}=0 
A_{2}x+B_{2}y+C_{2}z+D_{2}=0 
How to find the equation of a sheaf of planes.
If the line is

A_{1}x+B_{1}y+C_{1}z+D_{1}=0 
A_{2}x+B_{2}y+C_{2}z+D_{2}=0 
The equation of a sheaf of planes with axis, r, is:
λ(A_{1}x+B_{1}y+C_{1}z+D_{1})+μ(A_{2}x+B_{2}y+C_{2}z+D_{2})=0 
Dividing by λ and making k=μ/λ, the equation becomes:
(A_{1}x+B_{1}y+C_{1}z+D_{1})+k(A_{2}x+B_{2}y+C_{2}z+D_{2})=0
Find the equation of the plane that passes through the point (3,2,3) and belongs to the sheaf of planes with its axis on the following line:
Solution:
2x+3yz9+k(x+2y+3z+2)=0
6+6+39+k(3+49+2)=0, that is k=1
Therefore 2x+3yz9+(x+2y+3z+2)=0
The equation of the plane is:
x+5y+2z7=0