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`

Explanation

The equation of the plane through the intersection of the plane `x+y+z=1` and `2 x+3 y+4 z=5` is


`(x+y+z-1)+lambda(2 x+3 y+4 z-5)=0 `

`Rightarrow(2 lambda+1) x+(3 lambda+1) y+(4 lambda+1) z-(5 lambda+1)=0`


The direction ratios `a_{1}, b_{1}, c_{1}` of this plane are `(2 lambda+1),(3 lambda+1)` and `(4 lambda+1)`

The plane in equation (1) is perpendicular to `x-y+z=0`

Its direction ratios `{a}_{2}, {~b}_{2}, {c}_{2}` are `1,-1` and `1`

...