Equation of plane through the line of intersection of planes in Cartesian form is
For the standard equation of planes,
So, putting in equation (1), we have
x-2y + z-1 + λ(2x + y + z-8)=0
(1 + 2λ)x + (-2 + λ)y + (1 + λ)z-1-8λ=0 (2)
For plane the normal is perpendicular to line given parallel to this i.e.
Where A1, B1, C1 are direction ratios of plane and A2, B2, C2 are of line.
(1 + 2λ).1 + (-2 + λ).2 + (1 + λ).1=0
1 + 2λ-4 + 2λ + 1 + λ=0
-2 + 5λ=0
Putting the value of λ in equation (2)
9x-8y + 7z-21=0
9x-8y + 7z=21
For the equation of plane Ax + By + Cz=D and point (x1,y1,z1), a distance of a point from a plane can be calculated as
So, the required equation of the plane is 9x-8y + 7z=21, and distance of the plane from (1,1,1) is
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