Three two-candidate cells: a pivot with candidates {X,Y}, and two pincers sharing a house with the pivot, holding {X,Z} and {Y,Z}. Whichever value the pivot ends up being, one pincer is forced to Z.
A three-cell chaining pattern built entirely from bivalue cells (cells with exactly two candidates).
Find a bivalue cell (the pivot, candidates X and Y). Find two other bivalue cells that each share a house with the pivot - one holding {X,Z}, the other holding {Y,Z}, for some shared third digit Z.
If the pivot is X, the {X,Z} pincer can't be X, so it must be Z. If the pivot is Y, the {Y,Z} pincer must be Z instead. Either way, Z ends up placed at one of the two pincers - so Z can be eliminated from any cell that sees both pincers.
R2C4 has {3,7} (pivot), R2C9 has {3,5} (shares row 2), and R6C4 has {7,5} (shares column 4). Whichever digit R2C4 turns out to be, a 5 ends up at R2C9 or R6C4 - eliminate 5 from every cell that sees both.
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