For easy reference, the name xyz-wing is proposed to describe this technique without knowing whether the technique has already got a proper name.
Similar to the xy-wing, xyz-wing is a special case of forcing chains, and can be considered as another simplest application of such technique which involves 4 cells.
Similar to the xy-wing, the xyz-wing make use of the special property of a naked triple.
A naked triple is a group of three cells in the same unit that contain in total three candidates. Each cell can have two or three candidates. When this happens, the three candidates can be removed from all other cells in the same unit. Examples of naked triples are:
(123) (123) (123)
(123) (123) (12)
(123) (12) (23)
(12) (23) (13)
The second last case is the one we are interested in. We can formulate the special case of a naked triple where one cell contains 3 candidates and two cells contains only 2 candidates.
If three cells in the same unit contain candidates 'xy, xyz & xz' or 'xy, xyz & yz' or 'xz, xyz & yz' respectively, then x y and z can be removed from the candidates of all other cells in the unit.
Lets generalise the definition of the xyz-wing considering the first case 'xy, xyz & xz' (the other 2 cases are simliar):
If three cells a, b, c, with a and b in the same row or column, and b and c in the same box, with candidates xy, xyz and xz respectively, then x can be removed from the candidates of all cells (different from a,b,c) contained in the intersection of the two units that contain a, b and c as defined.
The reason for that is quite simple: a contains either x or y. If a contains x, then b contains yz and c contains xz; if a contains y, then b contains xz which forms a naked pair with the xz in cell c. Therefore either a, b or c contains x. Therefore a cell located inside the intersection of two units containing a, b and c cannot contain x.
An easy example would be:
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. . . | . . . | . . .
. xy----------------|-xyz . *
. . . | . . . | xz . .
----------+----------+----------
. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .
----------+----------+----------
. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .
Since one of r2c2, r2c7 & r3c7 must contain x, r2c9 cannot contain x.
Cells b & c need not be on the same column
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. . . | . . . | . . .
. xy----------------|-xyz . *
. . . | . . . | . \xz .
----------+----------+----------
. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .
----------+----------+----------
. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .
r2c2 is either x or y. If it's x, r2c9 cannot be x. If it's y, r2c7 is xz which forms a naked pair with r3c8, therefore r2c9 cannot be x. In either case, r2c9 cannot be x.
Let's try it with the following puzzle:
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6 . 3 | 1 . 8 | 9 4 2
. 8 . | . 3 . | 6 1 7
1 . 4 | . . . | 5 8 3
-------+-------+-------
7 . 1 | 3 . 6 | 8 9 5
9 6 8 | . . . | 3 2 .
. 3 5 | 8 . 9 | 7 6 .
-------+-------+-------
3 . 7 | . . . | 1 5 8
. 1 . | . 8 3 | . 7 6
8 . 6 | 7 . 1 | . 3 9
Having considered a hidden pair, 2 x-wings and colours on 2s, the candidates are reduced:
Amongst these cells, two xyz-wings can be identified enabling the puzzle to be completed. Can you see them?
Hint: 2 is removed in both cases.