[2200]

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The bidding is simple. LHO opens 15-17 1NT and RHO bids 3D invitational. LHO accepts.
Partner leads ♥Q. Plan your defence.

It seems that you have 2 lines:
1.declarer already has 9 tricks. ♥K, 6 diamonds, AQ of clubs finesse.
In this case, you need to overtake and hope to case 4 spades in a hurry;(pd may have Kxxx in spades but with QJ10x of hearts, the H lead looks safer and is normal)

2.even if diamonds are running, it's not enough.
In this case, play declarer for ♠K,♥K,diamonds AJ and ♣A. Parnter has QJ10xx in hearts, and all you need to do is to overtake with ♥A and play back a H.

Is it a guess?

[gszes]

trick 1 heart A
trick 2 !C K

If declarer has the !CAQ our play is irrelevant since they will score 1!H 2!C 6!D. If partner can come up with !CQ9x we just might be able to convince declarer to duck the !C K intending to take the finesse later after running the dia. If they duck we switch back to hearts and hope p can stand the stress of pitching on the diamonds. It is unrealistic to assume p will have more than 3 clubs given the lead so hearts is really our best shot overall we just have to convince declarer that finessing us for the Q is better than leading toward the spade K for their ninth trick. Returning the club T will only convince declarer to rise with the A hoping the suit blocks and lead toward their spade K (ick).

[FrancesHinden]

For partner to have Kxxx in spades, declarer's honour structure must be exactly (none), King, AJ, AQJ and declarer must be exactly 3-2, 3-3, or 2-2 in the majors as any 4-card spade holding blocks the suit.
For declarer to have only 8 tricks, declarer's honour structure must be K, K, A, A plus either one or both minor suit jacks (Axxx in diamonds in an AK AK 14-count might be possible), and declarer must have a doubleton heart.

There are a lot more ways to construct the second hand than the first. The first has a fixed honour structure and the possible shapes are 3244, 3235, 3253, 3343, 3334, 2245, 2254, 3236. I don't know if your partner would have come in over 1NT with Kxxxx QJ10xx xx x, which would eliminate some of these hands. The second has either 2 or 3 different honour combinations and slightly more flexibility on shape (4234, 4243, 3244, 3253, 3235, 4225 (maybe), 2245, 2254, 3236). Some of these are more or less likely relatively but I'm not up to that at the table.

So I would play for declarer to have only 8 tricks.

Finally, I would play the ace of hearts at trick one unless I was completely certain that declarer couldn't have K10xx K AJxx AQxx