= Solution
Put $D_1=X_1-Y_1$, $D_2=X_2-Y_2$ and
$$
B=\begin{pmatrix}1&0&-1&0\\0&1&0&-1\end{pmatrix}.
$$
The difference vector is $D=BZ$. Its mean is zero because the two sons have the same mean vector. Its <covariance matrix> is
$$
BVB^T=2\begin{pmatrix}a-c&b-c\\b-c&a-c\end{pmatrix}.
$$
For example, $\operatorname{Cov}(D_1,D_2)=b-c-c+b=2(b-c)$. Thus
$$
\boxed{\begin{pmatrix}D_1\\D_2\end{pmatrix}\sim N_2\!\left(\begin{pmatrix}0\\0\end{pmatrix},\ 2\begin{pmatrix}a-c&b-c\\b-c&a-c\end{pmatrix}\right).}
$$
The <positive definiteness> assumption supplies $a-b>0$ and $a+b-2c>0$: these are <eigenvalues> of the original <matrix> on within-son contrast and between-son difference directions. In particular the resulting difference <covariance> is positive definite.
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