Question

The distance of the point (–1, 2, –2) from the line of intersection of the planes 2x + 3y + 2z = 0 and x – 2y + z = 0 is :

(1) \(\cfrac{1}{\sqrt{2}}\)

(2) \(\cfrac{5}{2}\)

(3) \(\cfrac{\sqrt{42}}{2}\)

(4) \(\cfrac{\sqrt{34}}{2}\)

Answer 1

Correct answer is:  (4) \(\cfrac{\sqrt{34}}{2}\)
P₁ : 2x + 3y + 2z = 0

Direction vector of line L which is line of intersection of P₁ & P₂