Well, no, the cross product is <2, 0, 2>. There are no "x", "y", or "z" in it.

I don't know what you mean by "(x,y,z)= ". Given that <2, 0, 2> is a normal vector, the plane is of the form . If you choose (2, 0, 1) as your point, then 2(x- 2)+ 2(z- 1)= 0 which leads to 2x- 4+ 2z- 2= 0, 2x+ 2z= 6 and dividing by 2 gives your result.

It's easy to check it yourself, isn't it? (x,y,z)= (2, 0, 1) gives 2+ 1= 3, which is true. (1, 1, 2) gives 1+ 2= 3 which is true. (0, 0, 3) gives 0+ 3= 3, which is true. The three given points are on this plane and, geometrically, only one plane contains all of three independent points.(cartesian)

Is this right and how do I get the parametric form?

Another way you could have gotten this was to write Ax+ By+ Cz= D and put your given points into it: (2, 0, 1) gives 2A+ C= D. (1, 1, 2) gives A+ B+ 2C= D. (0, 0, 3) gives 3C= D. Multiplying the equation by a number does not change the plane so we can always choose any one constant and still have the same plane. In order to get an integer, take D= 3 so that C= 1. Now the first equation becomes 2A+ 1= 3 so A= 1. Finally, we have 1+ B+ 2= 3 and so B= 0. x+ z= 3 as you have.

As for getting the parametric equations, you can do this. z+ x= 3 means z= 3- x. y could be anything so we can let x and y be anything, say x= u, y= v, and then have z= 3- u. The parametric equations are x= u, y= v, z= 3- u.

A more direct way, without finding the Cartesian equation first, is to note that the vector from (2, 0, 1) to (1, 1, 2) is, as you say, (1- 2, 1- 0, 2- 1)= (-1, 1, 1), and the vector from (1, 1, 2) to (0, 0, 3) is (0- 1, 0- 1, 3- 2)= (-1, -1, 1). Since those vectors are independent, any vector in the plane can be written as a linear combination of then and so any point in the plane can be written in the form (x, y, z)= (2, 0, 1)+ s(-1, 1, 1)+ t(-1, -1, 1) which gives x= 2- s- t, y= s- t, z= 1+ s+ t.

You understand, I hope, that there are many choices for parametric equations of a given plane. These, x= u, y= v, z= 3- u, and x= 2- s- t, y= s- t, z= 1+ s+ t may look very different but note that z= 1+ s+ t= 3- (2- s- t)= 3- x. Taking u= 2- s- t= 2-(s+t) and v= s- t gives the same paratric equations as before.