Let A (1,1, 1), B (2, 3,5) and C (-1,0,2) be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is

Question

Let A (1,1, 1), B (2, 3,5) and C (-1,0,2) be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is (A) 2x - 3y + z + 2√14 = 0 (B) 2x -3 y + z -√14 = 0 (C) 2x - 3y + z + 2 = 0 (D) 2x - 3y + z - 2 = 0

Tardigrade Admin · Accepted Answer

A (1, 1, 1), B (2, 3, 5), C (-1 , 0, 2) direction ratios of AB are < 1 ,2 ,4 >. Direction ratios of AC are <-2,-1,1 >. Therefore, direction ratios of normal to plane ABC are < 2 ,-3,1 > As a result, equation of the plane ABC is 2x - 3y + z = 0. Let the equation of the required plane be 2x - 3y + z = k. Then |(k/√4+9+1)|=2 or k=±2√14 Hence, equation of the required plane is 2x - 3y + z + 2 √14 = 0

Q. Let $A (1, 1, 1), B (2, 3, 5)$ and $C (- 1, 0, 2)$ be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is

$2 x - 3 y + z + 214 = 0$

$2 x - 3 y + z - 14 = 0$

$2 x - 3 y + z + 2 = 0$

$2 x - 3 y + z - 2 = 0$

Q. Let A(1,1,1),B(2,3,5) and C(−1,0,2) be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is

2x−3y+z+214​=0

2x−3y+z−14​=0

2x−3y+z+2=0

2x−3y+z−2=0

Q. Let $A (1, 1, 1), B (2, 3, 5)$ and $C (- 1, 0, 2)$ be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is

$2 x - 3 y + z + 214 = 0$

$2 x - 3 y + z - 14 = 0$

$2 x - 3 y + z + 2 = 0$

$2 x - 3 y + z - 2 = 0$