AB=√[(4-2)²+(4-3)²+(1-1)²]=√(4+1+0)=√5
BC=√[(2-4)²+(7-4)²+(3-1)²]=√(4+9+4)=√17
CA=√[(2-2)²+(3-7)²+(1-3)²]=√(0+16+4)=√20
Perimeter=√5+√17+√20=3√5+√17.
Vector AB=(2,1,0)
Vector AC=(0,4,2)
Call vector AB vAB and vector AC vAC.
Dot product (vAB).(vAC)=(AB)(AC)cosBAC=√(5×20)cosBAC=10cosBAC.
Dot product=2×0+1×4+0×2=4, so 10cosBAC=4, cosBAC=0.4.
Area of ∆ABC=½(AB)(AC)sinBAC=5sinBAC=5√(1-0.4²)=5√0.84=√21=4.5826 square units approx.
The same result comes from Heron’s Formula using the perimeter:
Let s=(√5+√17+√20)/2=(3√5+√17)/2.
Area=√(s(s-√5)(s-√17)(s-√20))=√21.