What is the area of the triangle whose vertices are (0,0,0);(1,2,3)and (-3,-2,-1)?

Suppose ABC be a triangle with coordinates A(0, 0, 0); B(1, 2, 3) and C(-3, -2, -1).
Now AB, BC and AC are the sides of triangle ABC.
So; AB = $\sqrt{{\left(1-0\right)}^2+{\left(2-0\right)}^2+{\left(3-0\right)}^2} = \sqrt{14}$
BC = $\sqrt{{\left(-3-1\right)}^2+{\left(-2-2\right)}^2+{\left(-1-3\right)}^2} =4\sqrt{3}$
AC = $\sqrt{{\left(-3-0\right)}^2+{\left(-2-0\right)}^2+{\left(-1-0\right)}^2} = \sqrt{14}$
Using Heron's formula we have;
Area of triangle = $\frac{1}{4}\sqrt{\left(AB+BC+AC\right)\left(-AB+BC+AC\right)\left(AB-BC+AC\right)\left(AB+BC-AC\right)}$
$$\begin{gathered} =\frac{1}{4}\sqrt{\left(\sqrt{14}+4\sqrt{3}+\sqrt{14}\right)\left(-\sqrt{14}+4\sqrt{3}+\sqrt{14}\right)\left(\sqrt{14}-4\sqrt{3}+\sqrt{14}\right)\left(\sqrt{14}+4\sqrt{3}-\sqrt{14}\right)} \\ =\frac{1}{4}\sqrt{48\left(2\sqrt{14}+4\sqrt{3}\right)\left(2\sqrt{14}-4\sqrt{3}\right)} \\ =\frac{1}{4}\times 4\sqrt{3}\times \sqrt{{\left(2\sqrt{14}\right)}^2-{\left(4\sqrt{3}\right)}^2} \\ =\sqrt{3}\times \sqrt{56-48} \\ =2\sqrt{6} \end{gathered}$$