Calculate the unit normal vectors to the both sides of a plane Calculate the unit normal vectors to the both sides of a plane passing through three points with coordinates (1,0,1), (1,1,-1) and(-1,1,1).

My answer is [$\sqrt{6}/6 , \sqrt{6}/3 , \sqrt{6}/6$] and for the other side I just have to multiply by -1 which gives [$-\sqrt{6}/6 , -\sqrt{6}/3 , -\sqrt{6}/6$]

I would like to check whether I'm correct and for calculating the normal, does it matter which coordinate i choose to form the vectors??
Call $A=(1,0,1)$, $B=(1,1,-1)$, $C=(-1,1,1)$.

Each of the two vectors you found give $0$ in scalar product with $A-B$, $B-C$ (and $C-A$) and the modules of those vectors are $1$ in both cases (if you call the first $\vec{v}$, the second is $-\vec{v}$).

Without pen and paper next to me, I'd say you're right: you found two unit vectors, $\pm\vec{v}$, normal to the plane thru $A,B,C$.

P.S.: Changing system of coordinates may alter single components but NOT modules and orthogonality, which are indeed invariants.
Call $A=(1,0,1)$, $B=(1,1,-1)$, $C=(-1,1,1)$.

Each of the two vectors you found give $0$ in scalar product with $A-B$, $B-C$ (and $C-A$) and the modules of those vectors are $1$ in both cases (if you call the first $\vec{v}$, the second is $-\vec{v}$).

Without pen and paper next to me, I'd say you're right: you found two unit vectors, $\pm\vec{v}$, normal to the plane thru $A,B,C$.

P.S.: Changing system of coordinates may alter single components but NOT modules and orthogonality, which are indeed invariants.