Use cross product to find the sine of the angle between the vectors
u (2,3,-6)
v(2,3,6)
I'm not sure how you got this result ...
Use the formula I posted in my previous answer:
$\displaystyle
|\vec u \times \vec v|= |\vec u| \cdot |\vec v| \cdot \sin(\vec u, \vec v)
$
Plug in all the values you know:
$\displaystyle |(2,3,-6) \times (2,3,6)|= |(2,3,-6)| \cdot |(2,3,6)| \cdot \sin(\vec u, \vec v)$
$\displaystyle |(36, -24, 0)|= 7 \cdot 7 \cdot \sin(\vec u, \vec v)$
$\displaystyle 12 \cdot \sqrt{13}= 49 \cdot \sin(\vec u, \vec v)$
... and now solve for $\displaystyle \sin(\vec u, \vec v)$