= Solution
Apply <projective joins and meets by cross products> to each corresponding pair. The <cross products> of their representatives are proportional to $(0,1,0)$, $(0,0,1)$, and $(0,1,-1)$, respectively. Hence
$$
\boxed{PP':y=0,\qquad QQ':z=0,\qquad RR':y-z=0.}
$$
Their common point is
$$
\boxed{O=[1:0:0].}
$$
It is a valid <projective point>, because its representative is nonzero. The three equations force $y=z=0$, so this common <projective point> is unique. Thus the two <projective triangles> are in <perspectivity from a point> $O$.
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