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You could consider the position vector of $A$ to be $\overrightarrow{OA} = {\bf a}$.
You can then write $\overrightarrow{BA} = \overrightarrow{BO} + \overrightarrow{OA} = {\bf a} - {\bf b}$ etc.
Using $\overrightarrow{PS} = \overrightarrow{PA}+\overrightarrow{AS}$, can you find an expression for $\overrightarrow{PS}$?
For a regular tetrahedron we know that all of the edges have the same length, so $|AB|=|AC|$ etc. What extra facts about $PQRS$ might you be able to prove?