Question

The equation of the plane passing through (0, 2, 4) and containing the line (x + 3)/3 = (y – 1)/4 = (2 – z)/2 is 

(a) x – 2y + 4z – 12 = 0 

(b) 5x + y + 9z – 38 = 0 

(c) 10x – 12y – 9z + 60 = 0 

(d) 7x + 5y – 3z + 2 = 0

Answer 1

Answer is (c) 10x – 12y – 9z + 60 = 0

The equation of any plane through (0, 2, 4) is a(x – 0) + b(y – 2) + c(z – 4) = 0 which containing the line

So, 3a + 4b – 2c = 0

and –3a – b – 2c = 0

Solving, we get

a/–10 = b/12 = c/9

Hence, the equation of the plane is

–10x + 12(y – 2) + 9(z – 4) = 0 

–10x + 12y – 24 + 9z – 36 = 0 

10x – 12y – 9z + 60 = 0