I won't give a definitive answer in terms of ranking the three. Build 95% CIs around your parameters based on each, and if they're radically different, then your first step should be to dig deeper. Transform your data (though the LR will be invariant), regularize your likelihood, etc. In a pinch though, I would probably opt for the LR test and associated CI. A rough argument follows.
The LR is invariant under the choice of parametrization (e.g. T versus logit(T)). The Wald statistic assumes normality of (T - T0)/SE(T). If this fails, your CI is bad. The nice thing about the LR is that you don't need to find a transform f(T) to satisfy normality. The 95% CI based on T will be the same. Also, if your likelihood isn't quadratic, the Wald 95% CI, which is symmetric, can be kooky since it may prefer values with lower likelihood to those with higher likelihood.
Another way to think about the LR is that it's using more information, loosely speaking, from the likelihood function. The Wald is based on the MLE and the curvature of the likelihood at null. The Score is based on the slope at null and curvature at null. The LR evaluates the likelihood under the null, and the likelihood under the union of the null and alternative, and combines the two. If you're forced to pick one, this may be intuitively satisfying for picking the LR.
Keep in mind that there are other reasons, such as convenience or computational, to opt for the Wald or Score. The Wald is the simplest and, given a multivariate parameter, if you're testing for setting many individual ones to 0, there are convenient ways to approximate the likelihood. Or if you want to add a variable at a time from some set, you may not want to maximize the likelihood for each new model, and the implementation of Score tests offers some convenience here. The Wald and Score become attractive as your models and likelihood become unattractive. (But I don't think this is what you were questioning, since you have all three available ...)