= Solution
Let $A=R_{11}+2R_{12}+R_{22}$. Among the ten complete trios, six have two heterozygous parents: three have child $11$, two have child $12$ and one has child $22$. Their <conditional likelihood> contribution is
$$
\left(\frac{R_{11}}A\right)^3\left(\frac{2R_{12}}A\right)^2\frac{R_{22}}A=\frac{4R_{11}^3R_{12}^2R_{22}}{A^6}.
$$
Two more trios have a heterozygous parent and a $22$ parent, with child $12$ in each; they contribute $[R_{12}/(R_{12}+R_{22})]^2$. The remaining two parent pairs can produce only child $12$, so their conditional probabilities are one. Multiplying gives
$$
\boxed{L_{\rm exact}=\frac{4R_{11}^3R_{12}^4R_{22}}{(R_{11}+2R_{12}+R_{22})^6(R_{12}+R_{22})^2}.}
$$
The factor $4$ does not depend on the <genotype relative risks>, so the conventional proportional <likelihood function> is
$$
L\propto\frac{R_{11}^3R_{12}^4R_{22}}{(R_{11}+2R_{12}+R_{22})^6(R_{12}+R_{22})^2}.
$$
This is the expression in the original PDF, up to the parameter-independent constant. Its last denominator is $R_{12}+R_{22}$; the converted TeX's $R_{11}+R_{22}$ is a transcription error. The factor $4$ in the exact probability comes from the two heterozygous children of two heterozygous parents, each of which has two labelled transmission routes.
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