Find the equation of the plane through the line of intersection of the planes x - 3y + z + 6 = 0 and x + 2y + 3z + 5 = 0, and passing through the origin.

Equation of plane through the line of intersection of planes in Cartesian form is

(1)


For the standard equation of planes,



So, putting in equation (1), we have


x-3y + z + 6 + λ(x + 2y + 3z + 5)=0


(1 + λ)x + (-3 + 2λ)y + (1 + 3λ)z + 6 + 5λ=0 (2)


Now plane passes through (0,0,0) then it must satisfy the plane equation,


(1 + λ).0 + (-3 + 2λ).0 + (1 + 3λ).0 + 6 + 5λ=0


5λ=-6



Putting in equation (2)




-x-27y-13z=0


x + 27y + 13z=0


So, required equation of plane is x + 27y + 13z=0.


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