Is (2R,3S)-3,4-dimethylpentan-2-ol feasible to be dehydrated with POCl3 and pyridine? I was given a problem of reaction between (2R,3S)-3,4-dimethylpentan-2-ol and $\ce{POCl3}$ and pyridine. I know that $\ce{POCl3}$ will be attacked with hydroxyl group and the hydrogen on hydroxyl will be cleaved with pyridine so the hydroxyl group becomes good leaving group.

The reaction mechanism proceeds on E2 mechanism no matter whether the alcohol is primary, secondary, or tertiary. The problem is E2 mechanism can proceed if the stereochemistry of antiperiplanarity is fulfilled. When I drawn the stereochemistry of the alcohol, it turned that methyl on alpha-carbon and beta-carbon made the beta-hydrogen on these carbons cis to the hydroxyl group (the methyl on alpha carbon is wedge so the bonded hydrogen and hydroxyl group are dash), thus the reaction can not proceed.

Is that correct? If it is correct, will the reaction mechanism proceed on other mechanism?
I am uncertain whether or not you are assuming that the E2 product of a "cis elimination" is the (E)-alkene. Alternatively, you were given the geometry of the alkene and were asked to provide a mechanism for its formation. In either case, let's go through the steps.
You are correct that the alcohol 1 reacts with phosphorus oxychloride to form derivative 2a in addition to pyridine hydrochloride. Recall that there is free rotation about the C2-C3 bond in 2a, which is shown as its Newman projection in 2b in the anti-periplanar conformation ready for E2-elimination. Clearly, this elimination leads to (Z)-3,4-dimethylpent-2-ene. (Z)-4. If indeed it is the stereoisomer (E)-4 that is the target, then one can invoke SN2 displacement on 2a by chloride to invert the configuation at C2 leading to chloride 3a. Now E2-elimination of 3b leads to (E)-4.