Question

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

• A. 2x – 3y + z + $2\sqrt{14}$ = 0
• B. 2x – 3y + z – $\sqrt{14}$= 0
• C. 2x – 3y + z + 2 = 0
• D. 2x – 3y + z – 2 = 0

Answer 1

Answer: (A)

2x – 3y + z + $2\sqrt{14}$ = 0

Answer 2

A(1, 1, 1), B(2, 3, 5), C(–1, 0, 2) directions ratios of AB are (1, 2, 4) direction ratio of AC are ( – 2, – 1, 1)

\ direction ratios of normal to plane ABC are (2, – 3, 1)

\ equation of the plane ABC is 2x – 3y + z = 0

Let the equation of the required plane be

2x – 3y + z = k, then $\left| \frac{k}{\sqrt{4 + 9 + 1}} \right|$= 2

k = ±$2\sqrt{14}$

\ Equation of the required plane is 2x – 3y + z + $2\sqrt{14}$ = 0