Answer :

Given: Planes

∴ The equation of the planes passing through the line of intersection of these planes is given by;

2x − 3y + 4z − 1 + λ(x − y + 4) = 0

(2 + λ)x + (−3 − λ)y + (4)z − 1 + 4λ = 0

((2 + λ), (−3 − λ), (4)) ……(i)

Perpendicular plane is

In Cartesian form (2, −1, 1) ……(ii)

According to the given condition product of (i) and (ii) should be zero.

2(2 + λ) − (−3 − λ) + 4 = 0

∴ The required equation is

The given line x − 1 = 2y − 4 = 3z − 12

(1, 2, 4) is a point on the line

**∴** **the plane contains the line x − 1 = 2y − 4 = 3z − 12**

**OR**

Given: The point (1, 2, −4) and lines

The line perpendicular to these lines will be parallel to the line obtained by the cross product of these lines.

∴ The vector equation of the required line is

**∴** **The Cartesian equation of the required line is**

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