Question

The equation of the plane passing through the intersection of the planes 3x - y + 2z – 4 = 0 and x + y + z - 2 = 0 and passing through the point A(2, 2, 1) is given by 

A. 7x + 5y – 4z – 8 = 0 

B. 7x – 5y + 4z – 8 = 0 

C. 5x – 7y + 4z – 8 = 0 

D. 5x + 7y – 4z + 8 = 0

Answer 1

Given: Plane passes through the intersection of planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0. Point A(2, 2, 1) lies on the plane. 

To find: Equation of the plane. 

Formula Used: Equation of plane passing through the intersection of 2 planes P₁ and P₂ is given by P₁ + λP₂ = 0 

Explanation: 

Equation of plane is 

3x – y + 2z – 4 + λ (x + y + z - 2) = 0 … (1) 

Since A(2, 2, 1) lies on the plane, 

6 – 2 + 2 – 4 + λ (2 + 2 + 1 – 2) = 0 

2 + 3λ = 0

λ  = \(\frac{-2}3\)

Substituting in (1) and multiplying by 3, 

9x – 3y + 6z – 12 – 2 (x + y + z - 2) = 0 

9x – 3y + 6z – 12 – 2x – 2y – 2z + 4 = 0 

7x – 5y + 4z – 8 = 0 

Therefore the equation of the plane is 7x – 5y + 4z – 8 = 0