How to prove that in PG(2,R), if a projectivity takes 3 points to themselves, then the projectivity takes all points to themselves?

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If I have $(P,Q,R,X,Y,Z,...)\barwedge(P,Q,R,X',Y',Z',...)$ and all points lie on line $l$,

How do I go about proving $(X,Y,Z,...)=(X',Y',Z',...)$?

By a theorem, there exists a unique projectivity taking P,Q,R to itself.

If I define it by $f$, such that, $f(a)=a$,

Can I say, since there is no other projectivity apart from $f$ that takes P,Q,R to itself, in any tuple taking these points to themselves, the only possible projectivity is f. hence $f(X,Y,Z) = X', Y', Z'$

I can understand that it has to be true, but what is the correct way to prove this?