You are looking at two minor-suit losers, and clearly your best chance to reduce two losers to one loser is to draw two rounds of trump, retaining the 10, and lead a club to the 9.
If this rather deep finesse drives out the ace, you will have two winning clubs in your hand to discard two diamonds from dummy. If the ♣9 holds, continue with the ♣K to drive out the ace, again having two winning clubs if the king is taken. If both clubs win, enter your hand with the ♠10 and lead the ♣Q, discarding a diamond from dummy. The ♣J will then afford a second diamond discard. Making 4♠.
Can anything go wrong outside of the ♣9 losing to the 10? Yes, if East has five clubs headed by the ace and West has 10–x. Say East ducks the first club (best), wins the second club and leads a third club. Now if West started with three trumps and East two, you will not be able to discard two diamonds from dummy on your two high clubs, as West will ruff the third round of clubs. Down one, but unlikely.
What about drawing three rounds of trumps before leading a club to the 9? That works if East has the ace without the 10, takes the 9 and leads a diamond. No problem: Win the ♦K, cash the ♣K and come back to your hand with the ♦A to enjoy your ♣Q J. Not so fast: East might not be that generous and may duck both the ♣9 and the ♣K, leaving you stranded in dummy with only one hand entry, the ♦A, and no way to set up a club trick with the Q–J in your hand. Keep the faith. You are still alive. If diamonds break 3–3 (36%), play on diamonds after winning two club tricks. If they break 3–3, you will prevail. However, drawing two trumps rather than three gives you a much better chance of bringing this baby home.