= Solution
It is enough to prove the result for $0<\kappa<1$, since a construction for a smaller positive value gives the weaker vanishing requirement for any larger one. Apply part (b) with $\kappa/2$. For large $n$, part (c) gives linearly independent polynomials $P_0,Q_0$, and
$$
G(X,Y)=P_0(X)+YQ_0(X)
$$
satisfies $H(G)\leq C_1^n$ and has order at least
$$
\frac{(2-\kappa/2)n}{d}-2
$$
at $X=\alpha$ after setting $Y=\alpha$.
Set $\delta=\kappa/(4d)$. By the <rational multiplicity bound for a linear auxiliary polynomial>, once $q_1$ is sufficiently large, the one-variable polynomial $G(X,y)$ has multiplicity at most $\delta n+1$ at $p_1/q_1$. Consequently there is some integer $j\leq\delta n+1$ such that
$$
D_j^XG(p_1/q_1,y)\ne0.
$$
Put $F=D_j^XG$. The <normalized derivative of a polynomial> preserves integral coefficients and multiplies height by at most $2^n$, so $H(F)\leq C_2^n$. Differentiation lowers the vanishing order at $(\alpha,\alpha)$ by at most $j$; for sufficiently large $n$,
$$
\frac{(2-\kappa/2)n}{d}-2-j
\geq\frac{(2-\kappa)n}{d}.
$$
Finally write $F=P-YQ$ by replacing the coefficient of $Y$ by its negative. Then $F(p_1/q_1,y)\ne0$ and all the claimed bounds hold.
Solved by gpt-5.6-sol high.
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