= Solution
{c}
The four points form a <projective frame>: no three are collinear. By the allowed uniqueness theorem and <cross-ratio> invariance, use
$$
P_0=[1:0:0],\quad P_1=[0:1:0],\quad P_2=[0:0:1],\quad P_3=[1:1:1].
$$
Intersecting the indicated <projective lines> gives
$$
Q_0=[1:0:1],\qquad Q_1=[1:1:0],\qquad Q_2=[0:1:1].
$$
A point of $Q_0Q_2$ has coordinates $[a:b:a+b]$. Intersecting with $P_0P_1$, whose equation is $z=0$, gives $Q_3=[1:-1:0]$. On that line use its first two <homogeneous coordinates>. The <cross-ratio> from (a)(ii) is
$$
\boxed{(P_0P_1:Q_3Q_1)=\frac{(-1)(-1)}{(1)(-1)}=-1.}
$$
The <characteristic of a field> being zero ensures that $Q_3\ne Q_1$. The construction is the diagonal-point property of a <complete quadrangle>, producing <harmonic conjugate points>.
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