= Solution
Count one transmitted/untransmitted <allele> pair per observable parent. A homozygous parent contributes to $a$ or $d$; a heterozygous parent contributes to $b$ or $c$. When two heterozygotes have a heterozygous child, the parent-specific origin is ambiguous, but the combined contribution is unambiguously one to $b$ and one to $c$.
The derived contributions, ordered as $(a,b,c,d)$, are
$$
\begin{array}{c|rrrr}
\text{family}&a&b&c&d\\\hline
1&1&0&0&1\\2&0&2&0&0\\3&0&1&0&1\\4&0&1&1&0\\5&1&0&0&1\\6&0&2&0&0\\7&0&?&?&0\\8&0&0&2&0\\9&0&2&0&0\\10&0&1&0&0\\11&0&1&1&0\\12&0&1&0&1.
\end{array}
$$
In family 7, the observed father's transmission cannot be determined from the heterozygous child without the mother's genotype. In family 10, the homozygous child reveals the father's allele-1 transmission, but not the missing mother's complete transmitted/untransmitted pair.
Keeping only unambiguous observed-parent transmissions and omitting family 7 gives the raw table
$$
\boxed{(a,b,c,d)=(2,11,4,4).}
$$
This bookkeeping convention is not the same as a valid ordinary <transmission disequilibrium test> sampling rule. <Incomplete-trio transmission-selection bias> arises if only homozygous children make a missing-parent family usable. The usual complete-trio analysis excludes both families 7 and 10, giving
$$
\boxed{(a,b,c,d)_{\rm complete}=(2,10,4,4).}
$$
Neither assigning family 7 a half transmission each way nor inventing its mother's genotype is justified by the supplied data. A method using all twelve families must model or validly condition on the missing information.
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