Final answer:
To find the equation for the plane, vectors AB and AC were calculated from the given points, and then their cross product gave the normal vector of the plane, resulting in the final equation -2x + 6y - 7z = -3.
Step-by-step explanation:
To find an equation for the plane through the points (4,2,3), (-1,0,2), and (0,-1,1), we first need to find two vectors that are in the plane, which can be found by subtracting the coordinates of one point from another. Let's use vectors AB and AC where A=(4,2,3), B=(-1,0,2), and C=(0,-1,1).
Vector AB = B - A = (-1-4, 0-2, 2-3) = (-5, -2, -1)
Vector AC = C - A = (0-4, -1-2, 1-3) = (-4, -3, -2)
Now we can find a vector normal to the plane by taking the cross product of AB and AC:
AB × AC = |i j k|
|-5 -2 -1|
|-4 -3 -2|
This results in: i(4 - 6) - j(-10 + 4) + k(-15 + 8) = -2i + 6j - 7k
Therefore, the normal vector (n) is (-2, 6, -7). An equation of a plane can be expressed as n · (x, y, z) = d, where (x, y, z) are the coordinates of any point on the plane and d is a scalar. Plugging point A (4,2,3) into this equation to solve for d gives:
-2(4) + 6(2) - 7(3) = d => d = -3
The equation of the plane is therefore -2x + 6y - 7z = -3.