Triangle in Three Dimensions

Calculate the area of triangle $ABC$ with the following vertices: $A(1,3,0), B(0,2,5)$, and $C(\textrm{-}1,0,2)$.
Source: NCTM Mathematics Teacher, February 2006

SOLUTION
Apply the distance formula in three dimensions
$AB=\sqrt{(1-0)^2+(3-2)^2+(0-5)^2}$
$=\sqrt{1+1+25}$
$=\sqrt{27}$
$AC=\sqrt{(1+1)^2+(3-0)^2+(0-2)^2}$
$=\sqrt{4+9+4}$
$=\sqrt{17}$
$BC=\sqrt{(0+1)^2+(2-0)^2+(5-2)}^2$
$=\sqrt{1+4+9}$
$=\sqrt{14}$
Apply Heron’s formula for the area of a triangle $ABC$ with side lengths $a,b,c$
area =$\sqrt{s(s-a)(s-b)(s-c)}$ where $s=1/2(a+b+c)$
$s=1/2(\sqrt{27}+\sqrt{17}+\sqrt{14})$
$=6.53$
area =$\sqrt{6.53(6.53-\sqrt{27})(6.53-\sqrt{17})(6.53-\sqrt{14})}$
$=\sqrt{6.53(1.33)(2.41)(2.79)}$
$=7.64$

Answer: $7.64$ square units