We can begin by finding the direction vectors of two lines that lie in the plane containing the three given points.
One direction vector can be found by subtracting the coordinates of one of the given points from another given point. Let's subtract the coordinates of (-1, 3, -1) from (2, 7, -2):
(2-(-1), 7-3, -2-(-1)) = (3, 4, -1)
Another direction vector can be found by subtracting the coordinates of a different given point from the remaining given point. Let's subtract the coordinates of (-1, 3, -1) from (2, 8, 0):
(2-(-1), 8-3, 0-(-1)) = (3, 5, 1)
Now, we can find the cross product of these two direction vectors to obtain the normal vector of the plane:
(3, 4, -1) × (3, 5, 1) = (-1, -4, 3)
Let (x, y, z) represent a generic point on the plane. We know that the direction vector (9, 0, 0) is parallel to the plane. Therefore, a normal vector of the plane will satisfy the equation:
-1(x - (-1)) - 4(y - 3) + 3(z - (-1)) = 0
Simplifying the equation, we have:
-x + 1 - 4y + 12 + 3z - 3 = 0
-x - 4y + 3z + 10 = 0
Thus, an equation of the plane containing the three given points and with a coefficient of x equal to 9 is:
9(-x - 4y + 3z + 10) = 0
-9x - 36y + 27z + 90 = 0
Find an equation of a plane containing the three points (-1, 3, -1), (2, 7, -2), (2, 8, 0) in which the coefficient of x is 9.
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