Find the general form of the equation of the plane passing through the three points.[Be sure to reduce the coefficients in your answer to lowest terms by dividing out any common factor.] (3,1,6), (-6,-1,6), (3,5,-6)
A)2x - 9y - 3z + 21 = 0
B)2x - 9y - 3z = 0
C)3x + y + 6z = 0
D)2x - 9y - 3z - 21 = 0
E)3x + y + 6z - 252 = 0

Question

Quizplus · Accepted Answer

To find the equation of the plane passing through three non-collinear points (A,B,C), we can use the following formula:
(A-B) x (A-C) = n
where x denotes the cross product, and n is the normal vector of the plane.
Using the given points, we have:
(A-B) = (9,2,0)
(A-C) = (0,-4,-12)
(A-B) x (A-C) = (-24,-108,-38)
Dividing by the common factor of -2, we get:
n = (12,54,19)
The equation of the plane can then be written as:
12x - 54y - 19z + D = 0
To find the value of D, we can use any of the three points. Let's use A=(3,1,6):
12(3) - 54(1) - 19(6) + D = 0
D = 21
So the final equation of the plane is:
2x - 9y - 3z + 21 = 0
Therefore, the answer is choice A.

Find the General Form of the Equation of the Plane