Question -
Answer -
Given:
Equation of the planepasses through the intersection of the plane is given by
(3x – y + 2z – 4)+ λ (x + y + z – 2) = 0 and the plane passes through the points (2, 2, 1).
So, (3 × 2 – 2 + 2 × 1– 4) + λ (2 + 2 + 1 – 2) = 0
2 + 3λ = 0
3λ = -2
λ = -2/3 …. (1)
Upon simplification,the required equation of the plane is given as
(3x – y + 2z – 4) –2/3 (x + y + z – 2) = 0
(9x – 3y + 6z – 12 –2x – 2y – 2z + 4)/3 = 0
7x – 5y + 4z – 8 = 0
∴ The required equationof the plane is 7x – 5y + 4z – 8 = 0