Weak Beilinson conjecture for the transcendental motive of a hypergeometric K3 surface

Let $\alpha\in{\mathbb R}\setminus\{0,1\}$, let $X_\alpha$ be the associated smooth compactification, and let $\Delta_\alpha^+=\frac12(F_\infty+1)\Delta_\alpha$ be the real-Frobenius-invariant homology cycle. Assume that $\dim(H^{11})^{F_\infty=1}=0$, equivalently $\alpha>1$ by the lemma cited in the source. Let $h^2_{\mathrm{tr}}(X_\alpha)$ denote the transcendental part of the second motive. Weak Beilinson conjecture. There is an integral element $\xi\in K_3(X_\alpha)^{(3)}_{\mathbb Z}$ such that

$$\frac{1}{(2\pi\sqrt{-1})^2}\langle\operatorname{reg}_{\mathbb R}(\xi)\mid\Delta_\alpha^+\rangle\sim_{{\mathbb Q}^\times}L'(h^2_{\mathrm{tr}}(X_\alpha),0).$$

The source derives this as the expected weak Beilinson formula for the transcendental motive from the rank calculation and the regulator pairing, but does not establish the asserted integral element or the $L$-value formula.