the question
Let $ABCDA'B'C'D'$ be a parallelepiped, $O$ its centre and the points $M \in (AB), N \in (AD), P \in (A A')$ Denote by $Q, R, S$ symmetrical to the point $O$ of the points $M, N,$ respectively $P$. Knowing that $OM \cap (ADA')=\{Q\}, ON \cap (ABB') = \{R\}, OP \cap (ABC) =\{S\} $
a) prove that the points $Q, R, S$ are non-collinear;
b) prove that $(QRS) || (A'BD)$ and determine the ratio of the areas of the triangles $QRS$ and $A'BD$
my idea
for point a) I have been able to demonstrate that $D', A, Q$ are collinear by supposing they aren't collinear then saying that they all 3 $\in (AD'A), (D'AB)$, two different planes, which means they are collinear.
Similarly we get that $R, A,B'$ are collinear and $S, A,C$ are collinear.
I don't know what to do forward, I'm stuck. Hope one of you can help me! Thank you!