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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?