The 6n alternating binomial-sum divisibility conjecture

For positive integers $r$, $s$, $t$, and $n$, define

$$S_{6}(r,s,t;n)= \sum_{k=-n}^n(-1)^k{6n\choose 3n+k}^r{4n\choose 2n+k}^s{2n\choose n+k}^t.$$

The 6n divisibility conjecture. For all positive $r$, $s$, $t$, and $n$, $S_{6}(r,s,t;n)$ is divisible by both

$$2{6n\choose n}\qquad\text{and}\qquad 6{6n\choose 3n}.$$

This conjecture refines the preceding divisibility result, which gives divisibility by ${6n\choose n}$ and ${6n\choose 3n}$. The paper reports computational verification for $r+s+t\leq 10$ and $n\leq 100$, but no general proof is given.