Area of Triangle The position vectors of $A$ $B$ and $C$ relative to an origin $O$ are given by $OA=(2,1,3)$
   $OB=(0,-1,7)$ and $OC=(2,4,7)$
Part i) Show that angle $BAC= \cos^{-1}(\frac{1}{3})$
Part ii) Using the previous part calculate the exact area of the triangle $ABC$
I was able to do both parts easily. For part two I used the formula $A=\frac{1}{2}ab\sin c$ and got the answer $\frac{20 \sqrt3}{3}=11.547...$
Then just out of curiosity I decided to use Heron's formula to see if I would get the same answer. From Heron's formula I got $5\sqrt5=11.1803...$
My Question: Why do I get different answers?

Due to the Cosine Theorem:
$$\cos A = \frac{b^2+c^2-a^2}{2bc}$$
and since, through the Pythagorean Theorem:
$$c^2 = \|OA-OB\|^2 = 2^2+2^2+4^2 = 24,$$
$$b^2 = \|OA-OC\|^2 = 0^2+3^2+4^2 = 25,$$
$$a^2 = \|OB-OC\|^2 = 2^2+5^2+0^2 = 29,$$
we have:
$$\cos A = \frac{25+24-29}{10\sqrt{24}}=\frac{1}{\sqrt{6}}$$
so:
$$\sin A = \sqrt{\frac{5}{6}}$$
and:
$$ [ABC]=\frac{1}{2}bc\sin A = \frac{1}{2}5\sqrt{24}\sqrt{\frac{5}{6}}=5\sqrt{5}.$$
