If the lines x - 1 1 = y - 2 2 = z + 3 1 and x - a 2 = y + 2 3 = z - 3 1 intersects at the point P , then the distance of the point P from the plane z = a is :

Step-by-step solution

Let L 1 : x - 1 1 = y - 2 2 = z + 3 1 = λ Point on L 1 ≡ λ + 1 , 2 λ + 2 , λ - 3 And, L 2 : x - a 2 = y + 2 3 = z - 3 1 = μ Point on L 2 ≡ 2 μ + a , 3 μ - 2 , μ + 3 For finding P , we must have λ - 3 = μ + 3 ⇒ λ = μ + 6       . . . i 2 λ + 2 = 3 μ - 2 ⇒ 2 λ = 3 μ - 4       . . . ii Solving i   &   ii , we get λ = 22 and μ = 16 Therefore, P ≡ 23 , 46 , 19