Recently I had a curious discovery. Namely, I have made the following conjectures.

**Conjecture 1**. We have the identity
$$\sum_{k=0}^\infty\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k}=0.\label{1}\tag{1}$$

**Conjecture 2**. For any odd prime $p$ and positive integer $n$, we have
$$\frac1{(pn)^2}\bigg(\sum_{k=0}^{pn-1}\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k}-\left(\frac{-2}p\right)\sum_{k=0}^{n-1}\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k}\bigg)\in\mathbb Z_p,\label{2}\tag{2}$$
where $(\frac{\cdot}p)$ is the Legendre symbol and $\mathbb Z_p$ is the ring of $p$-adic integers.

**Conjecture 3**. Let $p$ be an odd prime. Then
$$\sum_{k=0}^{(p-1)/2}\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k}
\equiv\left(\frac{-2}p\right)\bigg(\frac98p^2q_p(2)^2-\frac32p\,q_p(2)-1\bigg)\ \ (\text{mod}\ p^3),\label{3}\tag{3}$$
where $q_p(2)$ denotes the Fermat quotient $(2^{p-1}-1)/p$. Also,
$$\sum_{k=0}^{p-1}\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k}
\equiv-\left(\frac{-2}p\right)+\frac{15}{16}p^2E_{p-3}\left(\frac14\right)
\ \ (\text{mod}\ p^3),\label{4}\tag{4}$$
where $E_{p-3}(x)$ denotes the Euler polynomial of degree $p-3$.

Note that the series in \eqref{1} has converging rate $27/32$. All the three conjectures can be easily checked numerically.

**QUESTION**。 How to prove the identity \eqref{1} and related $p$-adic congruences \eqref{2}, \eqref{3} and \eqref{4}? Is the WZ method helpful to solve the three conjectures?

Your comments are welcome!