area of a triangle - 2

Let the vertices of the triangle be $(10,0,0)$ , $(0,6,0)$ and $(0,0,8)$ . Then the area of the triangle is $\dfrac{1}{2}|(10\vec i -6\vec j) \times (8\vec k-6\vec j) |$ , where $\vec i, \vec j, \vec k$ are the unit vectors along the $X, Y$ and $Z$ -axes respectively. Carrying out the vector product and extracting the magnitude we get the area equal to $\dfrac{1}{2}\sqrt {12304}=2\sqrt {769}$ . Hence $a=2,b=769$ and $a+b=\boxed {771}$