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

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

$ \therefore (1+2\lambda)x + (1+3\lambda)y +(1+4\lambda)y$

$-5\lambda+1=0$ ---------(1)

The direction ratio a, b, c of the planes are

$(1+2\lambda), ) (1+3\lambda), (1+4\lambda)$

The above plane is perpendicular to $x-y+z=0$

The direction ratio of this plane is (1, -1, 1)

Since the two planes are perpendicular

$a_1a_2 + b_1b_2 +c_1c_2=0$

$ \therefore (1+2\lambda)(1)+(1+3\lambda)(-1) + (1+4\lambda)(1)=0$

$ \Rightarrow 1+2\lambda -1 - 3\lambda+1+4\lambda=0$

$3\lambda+1=0$

$ \Rightarrow \lambda = - \large\frac{1}{3}$