Find the vector equation of the plane through the line of intersection of the planes *x* + *y*+ *z* = 1 and 2*x* + 3*y* + 4*z* = 5 which is perpendicular to the plane *x* − *y* + *z* = 0.

#### Solution

`\text{ The equation of the plane passing through the line of intersection of the given planes is}`

\[x + y + z - 1 + \lambda \left( 2x + 3y + 4z - 5 \right) = 0 \]

\[\left( 1 + 2\lambda \right)x + \left( 1 + 3\lambda \right)y + \left( 1 + 4\lambda \right)z - 1 - 5\lambda = 0 . . . \left( 1 \right)\]

`\text{ This plane is perpendicular to x - y + z = 0 . So,}`

\[1 + 2\lambda - 1 \left( 1 + 3\lambda \right) + 1 + 4\lambda = 0 (\text{ Because a}_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]

\[ \Rightarrow 1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0\]

\[ \Rightarrow 3\lambda + 1 = 0\]

\[ \Rightarrow \lambda = \frac{- 1}{3}\]

`\text{ Substituting this in (1), we get }`

\[\left( 1 + 2 \left( \frac{- 1}{3} \right) \right)x + \left( 1 + 3 \left( \frac{- 1}{3} \right) \right)y + \left( 1 + 4 \left( \frac{- 1}{3} \right) \right)z - 1 - 5 \left( \frac{- 1}{3} \right) = 0\]

\[ \Rightarrow x - z + 2 = 0\]