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

Answer 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