3D Triangle Area

See @William Whitehouse for a nice solution using Heron's formula. Another easy way to do it is to construct two vectors using the three vertices, and then calculate the triangle area as half the magnitude of the cross product of those two vectors.

$\vec{v_1} = (-4,-5,-6) - (1,2,3) = (-5,-7,-9) \\ \vec{v_2} = (7,8,9) - (1,2,3) = (6,6,6) \\ \vec{v_1} \times \vec{v_2} = (12,-24,12) \\ \frac{1}{2} |\vec{v_1} \times \vec{v_2}| \approx 14.7$

Use Pythagoras to find the side lengths ( $\sqrt{155}, 6\sqrt{3}$ and $\sqrt{515}$ ) then apply herons formula (Area= $\sqrt{S(S-A)(S-B)(S-C)}$ where A B and C are the side lengths and S is the semi perimeter). This comes out as 14.69....