Solution:

Let’s say that the eq. of the plane that passes through the two-given planes
$x+y+z=1$ and $2 x+3 y+4 z=5$ is
$\begin{array}{l}
(x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 \\
(2 \lambda+1) x+(3 \lambda+1) y+(4 \lambda+1) z-1-5 \lambda=0 \ldots . .(1)
\end{array}$
Therefore the direction ratio of the plane is $(2 \lambda+1,3 \lambda+1,4 \lambda+1)$
And the direction ratio of another plane is $(1,-1,1)$
As, both the planes are $\perp$
On substituting in $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
$\begin{array}{l}
(2 \lambda+1 \times 1)+(3 \lambda+1 \times(-1))+(4 \lambda+1 \times 1)=0 \\
2 \lambda+1-3 \lambda-1+4 \lambda+1=0 \\
3 \lambda+1=0 \\
\lambda=-1 / 3
\end{array}$
Substituting the value of $\lambda$ in eq.(1) we obtain,
$\begin{array}{l}
\left(2 \frac{(-1)}{3}+1\right) x+\left(3 \frac{(-1)}{3}+1\right) y+\left(4 \frac{1}{3}\right) y=0 \\
\frac{1}{3} x-\frac{1}{3} z+\frac{2}{3}=0 \\
x-z+2=0
\end{array}$
As a result, the required equation of the plane is $x-z+2=0$