Why do we compute normal vector for computing projection? A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.27

(Exer 3.27) Consider the plane $H$ determined by the equation $x + y -z = 0$.
  
  
*
  
*What is a unit normal vector to $H$?
  
*Compute the image of $X:=H\cap \mathbb S^{2}$ under the stereographic projection $\Phi$.

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*
  
*For 2,  I computed $X$ to be a 3D circle, parametrised here, and its image to be $Y:= \Phi(X) = \{|z-(1+i)|^2 = 3\}$, but now I ask:
  
  
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For 1, What's the relevance of asking about the unit normal vector?

I computed the unit normal vectors to be $[1,1,-1]\frac{\pm 1}{\sqrt{3}}$. I observe their terminal points to be on the unit sphere.

Here are the parametrisations:
$Y:= \Phi(X) = \{|z-(1+i)|^2 = 3\}$ is parametrised:
$$\begin{bmatrix}
y_1(t)\\ 
y_2(t)\\ 
y_3(t)
\end{bmatrix} = \begin{bmatrix}
\sqrt{3}\cos(t) + 1\\ 
\sqrt{3}\sin(t) + 1\\ 
0
\end{bmatrix} = \begin{bmatrix}
1\\ 
1\\ 
0
\end{bmatrix} + \begin{bmatrix}
1\\ 
0\\ 
0
\end{bmatrix}\sqrt{3}\cos(t)+   \begin{bmatrix}
0\\ 
1\\ 
0
\end{bmatrix}\sqrt{3}\sin(t)$$
$X$ is parametrised:
$$\begin{bmatrix}
x_1(t)\\ 
x_2(t)\\ 
x_3(t)
\end{bmatrix} = \begin{bmatrix}
\sqrt{\frac 2 3} \cos[t]\\ 
-\sqrt{\frac 2 4} \sin[t] - \sqrt{\frac 2 {12}} \cos[t]\\ 
-\sqrt{\frac 2 4} \sin[t] + \sqrt{\frac 2 {12}} \cos[t]
\end{bmatrix} = \begin{bmatrix}
\sqrt{\frac 1 3}\\ 
-\sqrt{\frac 1 {12}}\\ 
\sqrt{\frac 1 {12}}
\end{bmatrix}\sqrt{2}\cos(t)+   \begin{bmatrix}
0\\ 
-\sqrt{\frac 1 {4}}\\ 
-\sqrt{\frac 1 {4}}
\end{bmatrix}\sqrt{2}\sin(t)$$
$$H = \{x + y -z = 0\} = \{[1,1,-1] \cdot [x,y,z]=0\} = \{[1,1,-1]\frac{1}{\sqrt{3}} \cdot [x,y,z]=0\}$$ is parametrised:
$$\begin{bmatrix}
h_1(r,s)\\ 
h_2(r,s)\\ 
h_3(r,s)
\end{bmatrix}=\begin{bmatrix}
1\\ 
0\\ 
1
\end{bmatrix}r +   \begin{bmatrix}
0\\ 
1\\ 
1
\end{bmatrix}s$$
 A: Here is one possible explanation. We are to find the image of $H\cap{\Bbb S}^2$ under $\Phi$. 
A solution may go like this: 
$$
Y\in\Phi(H\cap{\Bbb S}^2)\iff X=\Phi^{-1}(Y)\in H\cap{\Bbb S}^2 \subseteq H.
$$
The latter can be described by $ax_1+bx_2+cx_3+d=0$, where $(a,b,c)$ is a normal vector, giving an equation for $Y$ once the mapping $\Phi^{-1}$ is explicitly calculated.

EDIT: What I mean here is that $\Phi$ is the mapping
$$
\Phi\colon \underbrace{(x_1,x_2,x_3)}_{X}\in{\Bbb S}^2\to \underbrace{(y_1,y_2,y_3)}_{Y}=\left(\frac{x_1}{1-x_3},\frac{x_2}{1-x_3},0\right)\in{\Bbb R}^2.
$$
It is a bijection and
$$
\Phi^{-1}\colon(y_1,y_2,0)\to\Big(\underbrace{\frac{2y_1}{y_1^2+y_2^2+1}}_{x_1},\underbrace{\frac{2y_2}{y_1^2+y_2^2+1}}_{x_2},\underbrace{\frac{y_1^2+y_2^2-1}{y_1^2+y_2^2+1}}_{x_3}\Big).
$$
The necessary and sufficient condition for $Y$ to belong to the image is that $X$ (on the sphere) belongs to the plane $H$ (if a point is on the sphere, but not on $H$, then it does not belong to $H\cap{\Bbb S}^2$, hence, not mapped to the image due to one-to-one mapping), i.e. orthogonal to the normal vector
$$
x_1+x_2-x_3=0.
$$
When we substitute (and get rid of the denominator) we get the circle
$$
2y_1+2y_2-(y_1^2+y_2^2-1)=0\iff (y_1-1)^2+(y_2-1)^2=3.
$$
