In theory, if I have a certain function I can get his normal unit vector by using the gradient of it.
$$\hat{f} = \dfrac{\nabla f}{|| \nabla f ||}$$
Example (correction from answer):
$$ z = 2 -x -y$$ $$ f(x,y,z)= z + x + y -2 $$ $$ \nabla f(x,y,z)= \hat{i} + \hat{j} + \hat{k}$$ $$ \dfrac{\nabla z}{|| \nabla z ||}= \dfrac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})$$
Is that correct?
what about this example: $$ z = \sqrt{x^2+y^2} $$ $$ \nabla f(x,y,z)= \dfrac{x}{\sqrt{x^2+y^2}} \hat{i} + \dfrac{y}{\sqrt{x^2+y^2}} \hat{j} + -\hat{k}$$ $$ \dfrac{\nabla f}{|| \nabla f ||}= \dfrac{\dfrac{x}{\sqrt{x^2+y^2}} \hat{i} + \dfrac{y}{\sqrt{x^2+y^2}} \hat{j} + -\hat{k}}{\sqrt{ (\dfrac{x}{\sqrt{x^2+y^2}} )^2 + (\dfrac{y}{\sqrt{x^2+y^2}} )^2 + (-1)^2 }}$$
$$ \dfrac{\nabla f}{|| \nabla f ||}= \dfrac{\dfrac{x}{\sqrt{x^2+y^2}} \hat{i} + \dfrac{y}{\sqrt{x^2+y^2}} \hat{j} + -\hat{k}}{\sqrt{2}}$$
No, this is not correct. You must first write $f(x,y,z)=x+y+z-2$. Then, calculate $\nabla f=\vec i+\vec j+\vec k$ which gives $$\hat{f} = \dfrac{\nabla f}{|| \nabla f ||}=\frac{1}{\sqrt 3}(\vec i+\vec j+\vec k)$$ This is the required normal unit vector.