Find the equation of the plane through ...

Find the equation of the plane through the line of intersection of the planes `x + y + z = 1` and `2x + 3y + 4z = 5`which is perpendicular to the plane `x - y + z = 0`.

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To find the equation of the plane through the line of intersection of the planes \( x + y + z = 1 \) and \( 2x + 3y + 4z = 5 \) which is perpendicular to the plane \( x - y + z = 0 \), we can follow these steps: ### Step 1: Identify the equations of the given planes The equations of the two planes are: 1. \( P_1: x + y + z - 1 = 0 \) 2. \( P_2: 2x + 3y + 4z - 5 = 0 \) ### Step 2: Write the general equation of the plane through the line of intersection ...

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The equation of the plane passing through the line of intersection of the planes x+y+z=5 and 2 x + 3 y + 4z +5 = 0 and perpendicular to the plane x + y +z = 5 is

`x-z=10`

`x-z=20`

`x+y -2z=10`

`x+y -2z=20`

The equation of the plane passing through the line of intersection of the planes x + 2y = 3 and y - 2z + 1 =0 and perpendicular to the first plane is

`2x - y - 10 z = 9`

`2x - y - 9z = 10`

`2x - y + 10 z = 11`

`2x - y + 7z =11`

The equation of the plane passing through the line of intersection of the planes x + y + 2z = 1,2x + 3y + 4z = 7, and perpendicular to the plane x - 5y + 3z = 5 is given by:

A. x + 2y + 3z - 6=0

B. x + 2y + 3x + 6 = 0

C. 5x +7y +10z - 15=0

D. 3x + 4y + 5z +8=0