try this:
math.stackexchange.com/questions/2686606/equation-of-a-plane-passing-through-3-points
The points 𝐴(3,−1 ,2) ,𝐵(1 ,2 ,−4 ) and 𝐶 (−1 ,1 ,−2 ) are three vertices of a parallelogram ABCD.
(i) Show that the vector 𝑟 = −16𝑗 – 8𝑘 is perpendicular to the plan through 𝐴 ,𝐵 and 𝐶 .
(6 marks)
(ii) Find the cartesian equation of the plane through 𝐴 ,𝐵 and 𝐶 .
2 answers
vector BA = <2,-3,6>
vector BC = <-2, -1, 2>
a normal, or the cross-product would be <0,-2,-1> or <0, 2,1>
(I assume you know how to find the cross-product
since a normal is perpendicular to the plane and since
r = -16j - 8k is the same as the vector <0, -16, -8 >
which is a scalar multiple of <0, 2,1>
r must also be perpendicular to the plane.
equation:
the equation of the plane must take the form
0x + 2y + z = c
plug in any point on the plane, I will use (3,-1,2)
0 - 2 + 2 = c = 0
plane equation:
2y + z = 0
vector BC = <-2, -1, 2>
a normal, or the cross-product would be <0,-2,-1> or <0, 2,1>
(I assume you know how to find the cross-product
since a normal is perpendicular to the plane and since
r = -16j - 8k is the same as the vector <0, -16, -8 >
which is a scalar multiple of <0, 2,1>
r must also be perpendicular to the plane.
equation:
the equation of the plane must take the form
0x + 2y + z = c
plug in any point on the plane, I will use (3,-1,2)
0 - 2 + 2 = c = 0
plane equation:
2y + z = 0