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The equation of the plane passing through the point of intersection of the lines x - 1 3 = y - 2 1 = z - 3 2 , x - 3 1 = y - 1 2 = z - 2 3 and perpendicular to the line x - 2 2 = y - 3 3 = z - 2 1 is P = 0 . If the distance of the point 1,1 , 3 from P = 0 is k units, then the value of k 2 2 is equal to

Correct answer

7

Step-by-step solution

Any general point on the line x - 1 3 = y - 2 1 = z - 3 2 is 3 λ + 1 , λ + 2,2 λ + 3 For the point of intersection, this point must satisfy the line x - 3 1 = y - 1 2 = z - 2 3 ⇒ 3 λ - 2 1 = λ + 1 2 = 2 λ + 1 3 ⇒ λ = 1 So, point of intersection is 4,3 , 5 The required equation of plane is 2 x - 4 + 3 y - 3 + 1 z - 5 = 0 ⇒ 2 x + 3 y + z - 22 = 0 Hence, the distance of the plane from 1 , 1 , 3 = 2 + 3 + 3 - 22 14 = 14 = k

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