Write down an equation for the plane passing through the points [2,0,0][2,0,0], [−3,3,−2][−3,3,−2], and [2,−1,1][2,−1,1]. For example: x+y+z=1.

The equation of a plane in 3D space is given by the formula:

A(x - x1) + B(y - y1) + C(z - z1) = 0

where (x1, y1, z1) is a point on the plane and (A, B, C) is the normal vector to the plane.

The normal vector can be found by taking the cross product of two vectors that lie in the plane. We can get these two vectors by subtracting the coordinates of the given points:

Vector1 = [2,0,0] - [-3,3,-2] = [5,-3,2] Vector2 = [2,0,0] - [2,-1,1] = [0,1,-1]

Now, we calculate the cross product of these two vectors to get the normal vector:

Normal vector = Vector1 x Vector2 = [(-3*-1) - (21), (20) - (5*-1), (51) - (-30)] = [3, 5, 5]

So, the equation of the plane is:

3(x - 2) + 5(y - 0) + 5(z - 0) = 0

Simplifying this gives:

3x + 5y + 5z = 6

So, the equation of the plane passing through the points [2,0,0], [-3,3,-2], and [2,-1,1] is 3x + 5y + 5z = 6.