Question

Consider the lines

L₁: (x + 1)/3 = (y + 2)/1 = (z + 1)/2

L₂: (x – 2)/1 = (y + 2)/2 = (z – 3)/3

The distance of the point (1, 1, 1) from the plane passing through the point (–1, – 2, –1) and whose normal is perpendicular to both the lines L₁ and L₂ is

Answer 1

Answer is (c) 13/√75

The equation of any plane passing through (– 1, –2, –1) is

a(x + 1) + b(y + 2) + c(z + 1) = 0 ...(i)

which is perpendicular to the lines

Hence, the equation of the plane is

(x + 1) + 7(y + 2) – 5(z + 1) = 0 

x + 7y – 5z + 10 = 0.

Therefore, the required distance