Let $X$ follow Binomial distribution with parameters $12$ and $p$, let $q = 1 - p$. If
\[
\sum_{x=0}^{12}(x - 12p)^2 \cdot {}^{12}C_x \cdot q^{12 - x} \cdot p^x = \frac{8}{3} \quad \text{and} \quad P(X>10) = \left(\frac{2}{3}\right)^K, \, (K>1),
\]
then $K =$