Question

find the equation of the plane that passes through the line of intersection of the planes x - z = 10 and y + 5z = 9, and is perpendicular to the plane 3x + 2y - 2z = 7.

Answer 1

see the image to verify the sum

Answer 2

Given that the plane required passes through the line of intersection of the planes x - z = 10 and y + 5z = 9,

Hence equation of plane would be of the form

x-z-10+k(y+5z-9) =0

Regrouping we have

x+ky+z(5k-1)-10-9k =0

The normal of this plane will have direction ratios as coefficients of x,y,z

i.e. (1, k, 5k-1)

The perpendicular plane has normal as (3,2,-2)

Since the two normals are also perpendicular the dot product =0

3+2k-10k+2 =0

-8k+5=0

k =5/8

Substitute in the equation of the plane to get

x+ky+z(5k-1)-10-9k =0 equals

x+5y/8+z{(25/8)-1}-10-9(5/8) =0

8x+5y+17z -125 =0 is the required plane.