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Find an equation of the plane that passes through (3, 2, 1), (1, 2, 3) and (9, 5, 4).

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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

User Bhavani Kannan
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