Official solution
Given,
P1 : x + y + z = 1
P1 : 2x
+ 3y + 4z = 5
Equation of the plane passing through the line
of intersection of the plane P1 and P2 is :
P1 + \[\lambda \]P2 = 0
\[ \Rightarrow \] (x + y + z --1) + \[\lambda (2x + 3y + 4z -- 5) = 0
\[ \Rightarrow \] x(1 + 2\[\lambda \]) + y(1 + 3\[\lambda \]) + z(1 + 4\[\lambda \]) - 5\[\lambda \] - 1 = 0 .....(1)
Direction Ratio (D.R) of this plane = (1 + 2\[\lambda \], 1 + 3\[\lambda \], 1 + 4\[\lambda \])
Plane (1) is perpendicular to x - y + z = 0, whose D.R = (1, -1, 1)
As they are perpendicular so dot product of D.R = 0
\[ \therefore \] (1) (1 + 2\[\lambda \]) + (--1) (1 + 3\[\lambda \]) + (1) (1 + 4\[\lambda \]) = 0
\[ \Rightarrow \] 1 + 2\[\lambda \] --1 -- 3\[\lambda \] + 1 + 4\[\lambda \] = 0
\[ \Rightarrow \] \[\lambda \] = \[ - {1 \over 3}\]
Putting the value of \[\lambda \] in equation (1), we get
\[ \Rightarrow \] \[{x \over 3} - {z \over 3} + {2 \over 3} = 0\]
\[ \Rightarrow \] x - z + 2 = 0
Vector form of this plane,
\[\mathop r\limits^ \to . \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } \right) + 2 = 0\]