= Solution
The corresponding side equations, computed using <projective joins and meets by cross products>, are
$$
\begin{aligned}
PQ&:x-y-2z=0,&P'Q'&:x-2y-z=0,\\
QR&:x-y+z=0,&Q'R'&:x-2y+z=0,\\
RP&:x+2y-2z=0,&R'P'&:x-z=0.
\end{aligned}
$$
Subtracting the two equations in each row and substituting back gives the three <projective points>
$$
\boxed{X=[3:1:1],\qquad Y=[1:0:-1],\qquad Z=[2:1:2].}
$$
For example, the first pair forces $y=z$ and $x=3z$; the second pair forces $y=0$ and $x=-z$; the last pair forces $x=z=2y$. All three satisfy
$$
\boxed{x-4y+z=0,}
$$
since $3-4+1=0$, $1-0-1=0$, and $2-4+2=0$. The coefficient vector is nonzero, so this is a <projective line>. Thus \b[the corresponding-side intersections are <collinear>], explicitly confirming <Desargues's theorem>.
The diagram uses the <affine chart> $x+y+z\ne0$ with coordinates $u=y/(x+y+z)$ and $v=z/(x+y+z)$. Here the center $O$ is finite, the axis is $u=1/5$, and $Y$ lies on the <projective line at infinity>. The apparent parallelism of $QR$ and $Q'R'$ records their common vertical point at infinity, which also belongs to the axis.
\Image[/past-exam-of-the-short-options-course-of-the-university-of-oxford/2017-desargues-configuration.png]
{title=Desargues's theorem: three connectors meet at O and three side intersections lie on one projective line}
{description=Original drawing in the affine chart x+y+z=1. Blue and orange triangles are in perspectivity from O. Two side intersections X and Z lie on the red vertical axis; the third intersection Y is its vertical point at infinity.}
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