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. Also find the distance of the plane obtained above, from the origin.

Equation of given planes are
P1 ➡️x+y+z=0
P2 ➡️2x+3y+4z=0
Equation of plane through the line of intersection of planes P1, P2 is
P1 + λP2 = 0
(x+y+z) + λ(2x+3y+4z) = 0
(1+2λ)x + (1+3λ)y + (1+4λ)z + ( -λ) = 0 — (1)
Given that plane represented by equation (1) is perpendicular to plane x-y+z=0
So we use formula a1a2+b1b2+c1c2=0
So,(1+2λ).1 + (1+3λ)(-1) + (1+4λ).1 = 0
1+2λ -1-3λ +1+4λ = 0
3λ+1 = 0
λ = -1/3
Put λ = -1/3 in equation (1), we get
(1-2/3)x + (1-1)y + (1-4/3)z -1/3 = 0
x/3 - z/3 - 1/3 = 0
x - z + 2 = 0
General points on the line:
x=2+3λ, y=-1+4λ, z=2+2λ
The equation of plane: ➡️r.(i-k)=0
The point of intersection of the line and the plane :
Substituting general point of the line in the equation of plane and finding the particular value of λ.
[(2+3λ)i + (-1+4λ)j + (2+2λ)k].(i-k) = 0
➡️(2+3λ).1 + (-1+4λ)(-1) + (2+2λ).(-1) = 0
➡️12 + λ = 0 or λ = -4
The point of intersection is :(2+3(-4), -1+4(-4), 2+2(-4)) = (-10,-17,-6)
Distance of this point from (0,0,0) is = √((-10)²+(-17)²+(-6)²) = √(100+289+36) = √425 units.