Area of a triangle using vectors I have to find the area of a triangle whose vertices have coordinates 
O$(0,0,0)$, A$(1,-5,-7)$ and B$(10,10,5)$
I thought that perhaps I should use the dot product to find the angle between the lines $\vec{OA}$ and $\vec{OB}$ and use this angle in the formula:
area $= \frac{1}{2}ab\sin{C}$
These are my steps for doing this:
$\mathbf{a} \cdot \mathbf{b} = \begin{vmatrix} {\mathbf{a}} \end{vmatrix}\begin{vmatrix} {\mathbf{b}} \end{vmatrix} \sin{\theta} $ 
Let $\mathbf{a} = \begin{pmatrix} 1 \\ -5 \\ -7 \end{pmatrix}$ and let $\mathbf{b} = \begin{pmatrix} 10 \\ 10 \\ 5 \end{pmatrix}$
$\therefore \begin{pmatrix} 1 \\ -5 \\ -7 \end{pmatrix} \cdot \begin{pmatrix} 10 \\ 10 \\ 5 \end{pmatrix} = (5\sqrt{3})(15)\sin{\theta} $
$\therefore \sin{\theta} = -\dfrac{1}{\sqrt{3}}$
If I substitute these values into the general formula:
area $= \frac{1}{2}ab\sin{C}$
I get:
area $= \frac{1}{2}(5\sqrt{3})(15)(-\dfrac{1}{\sqrt{3}})$
$\therefore$ area $= -\dfrac{75}{2}$
However this isn't right, the area should be $\dfrac{75}{\sqrt{2}}$
I feel I'm missing something really obvious but I can't spot it, can anyone help? 
Thank you. 

Alternative solution
$a = |OA| = \sqrt{1^2 + 5^2 + 7^2} = \sqrt{75} = 3 \sqrt{5}$
$b = |AB| = \sqrt{(10-1)^2 + (10--5)^2 + (5--7)^2} = \sqrt{450} = 15\sqrt{2}$
$c = |BO| = \sqrt{10^2+10^2+5^2} = 15.$
Now you can calculate the semiperimeter $s$ which is just $\frac{1}{2} \left(3 \sqrt{5} + 15 \sqrt2 + 15 \right)$, and use Heron's formula to find the area:
$$A = \sqrt{s(s-a)(s-b)(s-c)}$$
