find an equation for the plane passing through the points Pi(3,1-2), Pii(-1,2,4) and Piii(2,-1,1)

First you will need 2 vector on the plane, I took the vector between the first and third point, and the first and second point and got direction

vectors (-1,-2,3) and (-4,1,6)

We now need a normal to these two vectors.
You should have been taught a method to do this, assuming you know how to do this, you should get (5,2,3).
That is, the dot product between (5,2,3) and each of the above direction vectors is zero. I checked for that.

So the equation of our plane is
5x + 2y + 3z = k
sub in any one of the original points, I will use (3,1,-2), we get
15 + 2 - 6 = k
k = 11

So the equation of the plane is
5x + 2y + 3z = 11

To find an equation for the plane passing through three non-collinear points, we can use the cross product of two vectors that lie on the plane. Let's call the three points Pi(3,1,-2), Pii(-1,2,4), and Piii(2,-1,1).

1. Find two vectors on the plane:
Let's find two vectors by subtracting the coordinates of one point from the other two points.

Vector A = Pii - Pi = (-1, 2, 4) - (3, 1, -2) = (-4, 1, 6)
Vector B = Piii - Pi = (2, -1, 1) - (3, 1, -2) = (-1, -2, 3)

2. Find the cross product of vectors A and B:
The cross product of the two vectors will give us a vector that is orthogonal to the plane.

Cross product = A x B = (A2B3 - A3B2, A3B1 - A1B3, A1B2 - A2B1)
= ((1 * 3) - (6 * -2), (6 * -1) - (-4 * 3), (-4 * -2) - (1 * -1))
= (15, -14, -3)

3. Write the equation of the plane:
Now, we have a normal vector (orthogonal to the plane) and a point on the plane (Pi). The equation of the plane is given by:

Ax + By + Cz = D

where A, B, C are the components of the normal vector, and D can be found by substituting the coordinates of a point on the plane into the equation.

Substituting the values, we get:

15x - 14y - 3z = D

You can keep D as it is or simplify it further based on your requirements.