I play a spade at trick 2, to the K.
If it wins, I play a spade to the 10 next.. if I can get 2 spade tricks while losing only 1, I have lots of chances, including 3-3 spades or the club K onside. Exactly how I combine those chances depends on the defence
If the spade K loses to the Ace, then my line again depends on the defence, but I am going to assume that the club K is onside... giving me 11 top tricks and I will already have lost a trick so the count is rectified.
Say they return a spade. I pop the Q... I do not hook the 10. I then run the hearts.
I hope the 4 card ending has me with 10 void Ax x and dummy with void void Q AQx and that I can read the end position.
Give LHO the spade J and the club K, and he has to reduce to J void void Kxx while RHO has to keep 2 diamonds, so can't hold clubs. Now when I cash the diamond A, LHO is crushed.
If RHO has the spade J, then in the 4 card ending, he has to keep that card and 2 diamonds and only one club, while LHO holds Kxx of clubs and an immaterial 4th card.
Now, the club hook and cashing the club Ace crushes RHO.
There are other variants depending on the defence. Note that my constructions suggest a brilliant defence available to RHO should he win trick 2 with the spade Ace: he can and maybe should switch to a club into the AQ... obviously, not if he holds the K
'one of the great markers of the advance of human kindness is the howls you will hear from the Men of God' Johann Hari