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:

First, find the line of intersection of the two given planes:
1. Write the equations of the planes in the form Ax + By + Cz = D.
- The first plane: x + y + z = 1
- The second plane: 2x + 3y + 4z = 5
2. Set up a system of equations with the two plane equations and solve for x, y, and z.
- Multiply the first plane equation by 2 and subtract it from the second plane equation:
(2x + 3y + 4z) - 2(x + y + z) = 5 - 2(1)
-x + y + 2z = 3
- Now, we have a system of two equations:
x + y + z = 1 (Equation 1)
-x + y + 2z = 3 (Equation 2)
3. Solve this system of equations to find the values of x, y, and z.
- Add Equation 1 and Equation 2:
(x + y + z) + (-x + y + 2z) = 1 + 3
2y + 3z = 4
- Let y = t (a parameter), then we can express z in terms of t: z = (4 - 2t)/3
- Substitute the values of y and z into Equation 1 to solve for x:
x + t + (4 - 2t)/3 = 1
3x + 3t + 4 - 2t = 3
3x + t + 4 = 3
3x + t = -1
x = (-1 - t)/3
4. The line of intersection can be parameterized as:
x = (-1 - t)/3
y = t
z = (4 - 2t)/3

Next, find the direction vector of the line of intersection:
5. The direction vector of the line of intersection can be obtained by taking the coefficients of x, y, and z from the parameterized equations:
Direction vector = (1/3, 1, -2/3)

Finally, find the equation of the plane perpendicular to the plane x + y + z = 0:
6. The normal vector of the plane x + y + z = 0 is (1, 1, 1).
7. The direction vector of the line of intersection is parallel to the plane we are looking for, so the normal vector of the plane we want is perpendicular to the direction vector of the line of intersection.
8. Use the cross product to find the normal vector of the plane we want:
Normal vector of the plane = (1, 1, 1) x (1/3, 1, -2/3)
= (1*(-2/3) - (1/3)*1, (-2/3