Any plane through the line of intersection of the planes x + y + z – 1 = 0 and 2x + 3y + 4z – 5 = 0 is (x + y + z – 1) + k( 2x + 3y + 4z – 5) = 0 i.e. (1 + 2k) x + (1 + 3k) y + (1 + 4k) z – (1 + 5k) = 0 ...(1) Since it is perpendicular to the plane x – y + z = 0 ...(2) ∴ Their normals are perpendicular ⇒ (1 + 2k) (1) + (1 + 3k) (–1) + (1 + 4k) (1) = 0⇒ 1 + 2k – 1 – 3k + 1 + 4k = 0$\Rightarrow 3k=-1 \Rightarrow k=-\frac{1}{3}$Putting in (1), we get, $\left(1-\frac{2}{3}\right)x$ $+(1-1)y+\left(1-\frac{4}{3}\right)z-\left(1-\frac{5}{3}\right)=0$$\Rightarrow \frac{1}{3}x - \frac{1}{3}z + \frac{2}{3}=0$⇒ x – z + 2 = 0, which is the required equation