Given: Point A(2, 3, 4) lies on a plane which is parallel to 5x – 6y + 7z = 3

To find: Equation of the plane

Formula Used: Equation of a plane is

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

where a:b:c is the direction ratios of the normal to the plane

(x₁, y₁, z₁) is a point on the plane.

Explanation:

Since the plane (say P₁) is parallel to the plane 5x – 6y + 7z = 3 (say P₂), the direction ratios of the normal to P₁ is same as the direction ratios of the normal to P₂.

i.e., direction ratios of P₁ is 5 : -6 : 7

Let the equation of the required plane be

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

Here a = 5, b = -6 and c = 7

Since (2, 3, 4) lies on the plane,

5(x - 2) – 6 (y - 3) + 7 (z - 4) = 0

5x – 6y + 7z – 10 + 18 – 28 = 0

5x – 6y + 7z = 20

The equation of the plane is 5x – 6y + 7z = 20