Let A=(3, -1, 2), B=(1, -1, 2), C=(4, -2, 1).
Then find the distance between the vertices of triangle:
c = AB = (3−1)2+(−1−(−1))2+(2−2)2=2,
a = BC = (1−4)2+(−1−(−2))2+(2−1)2=11,
b =AC = (3−4)2+(−1−(−2)2+(2−1)2)=3 .
The last step is to use the Heron's formula for the area of triangle:
A=2(a+b+c)2(b+c−a)2(a−b+c)2(a+b−c)==41(2+11+3)⋅(11+3−2)⋅(2−11+3)⋅(2+11−3)==41(7+43−11)⋅⋅(211+11−33+23+33−3−4−211+23)==41(43−4)(4+43)=41⋅4(3−1)(3+1)=2.
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