Question

The equation of the plane passing through the points (1, −2, 2) and (−3, 1, −2) and perpendicular to the plane 2x + y − z + 6 = 0 is

(a) x - 12y - 10z - 5 = 0 

(b) x - 12y - 10z + 5 = 0

(c) x - 12y - 10z -10 = 0

(d) x - 12y - 10z + 10 = 0 

Answer 1

Let E ≡ ax + by + cz + d = 0 be the required plane. It passes through (1, −2, 2) and (−3, 1, −2). This implies

a - 2b + 2c + d = 0 ...(1)

and -3a + b - 2c + d = 0 ...(2)

E = 0 is perpendicular to the plane 2x + y − z + 6 = 0. This gives

2a + b - c = 0 ...(3)

Subtracting Eq. (2) from Eq. (1), we get

4a - 3b + 4c = 0 ...(4)

From Eqs. (3) and (4), we get

12a + b = 0 

b = - 12a

Therefore, 

c = 2a + b = - 10a

and d = - a + 2b - 2c = - a + 2(-12a) + 20a = - 5a

Hence, 

E ≡ ax + by + cz + d ≡ ax - 12ay - 10az - 5a = 0

E ≡ x - 12y - 10z - 5 = 0