Final answer:
To find an equation of the plane that passes through the given points (3, 2, 1), (1, 2, 3), and (9, 5, 4), we can use the formula for a plane and the cross product of two vectors.
Step-by-step explanation:
To find an equation of the plane that passes through the given points (3, 2, 1), (1, 2, 3), and (9, 5, 4), we can use the formula for a plane:
Ax + By + Cz = D
where (A, B, C) is a normal vector to the plane and D is a constant. We can start by finding two vectors that lie in the plane by subtracting one point from another:
(1, 2, 3) - (3, 2, 1) = (-2, 0, 2)
(9, 5, 4) - (3, 2, 1) = (6, 3, 3)
Now, we can find a normal vector to the plane by taking the cross product of these two vectors:
N = (-2, 0, 2) x (6, 3, 3)
Once we have the normal vector, we can use one of the given points to solve for D:
3A + 2B + C = D
Substituting the values of A, B, and C from the normal vector, we have:
3(-2) + 2(0) + 2C = D
Simplifying, we get:
D = -6
Therefore, the equation of the plane is:
-2x + 2z - 6 = 0