Finding the points that line on a plane

Question

Let $P$ denote the plane given by the point-normal equation: $0 = (1,2,−1)·((x,y,z)−(1,1,1))$

How do I find the points $(x, y, z)$ that lie on the plane $P$?

Answer 1

You must write better your equation:

$0 = (1,2,−1)·((x,y,z)−(1,1,1))$

Implies

$0=(1,2,-1)(x-1,y-1,z-1)$

So

$(x-1)+2(y-1)-z+1=0$

Then

$x+2y-z=2$

$P=\{(x,y,z): x+2y-z=2\}$

Answer 2

The points on the plane satisfy the given equation. I think this is just an exercise in using the dot product. Recall that for vectors $a,b,c,$ we have that $$a\cdot(b+c)=a\cdot b + a\cdot c,$$ so that your equation simplifies to $$0=(1,2,-1)\cdot(x,y,z)-(1,2,-1)\cdot(1,1,1).$$

You should be able to complete this. Calculate the products and simplify the equation to get a nonhomogeneous linear equation in $x,y,z$