A(1, 1, 0) B(2, 1, 1) and C(1, 1, 2)


The magnitude of the cross product of two vectors
is the area of the parallelogram which has adjacent
sides equal to the magnitudes of the two vectors.

We form the vectors AB and AC

AB = < 2-1, 1-1, 1-0 > = < 1, 0, 1 >

AC = < 1-1, 1-1, 2-0 > = < 0, 0, 2 >

We find their cross product AB×AC

 = [(0)(2)-(1)(0)]i - [(1)(2)-(1)(0)]j + [(1)(0)-(0)(0)]k
 
= 0i - (2-0)j + 0k = < 0, -2, 0 >

The magnitude of < 0, -2, 0 > is
 
√0² + (-2)² + 0² = √0 + 4 + 0 = √4 = 2

The triangle's area is  the area of that parallelogram.

Answer = (2) = 1

Edwin