Three Equations, Three Variables!

Note that $x,y,z\ne0$ . Then we can write equivalently the system as follows

$\quad\;\left\{\begin{array}{l}x(x^2+y^2+z^2)-2xyz=2\\y(x^2+y^2+z^2)-2xyz=30\\z(x^2+y^2+z^2)-2xyz=16\end{array}\right.$

$\Leftrightarrow \left\{\begin{array}{l}x(x^2+y^2+z^2)-2xyz=2\\(y-z)(x^2+y^2+z^2)=14\\(z-x)(x^2+y^2+z^2)=14\end{array}\right.$

$\Leftrightarrow \left\{\begin{array}{l}x(x^2+y^2+z^2)-2xyz=2\\(y-z)(x^2+y^2+z^2)=14\\y=2z-x\end{array}\right.$

$\Leftrightarrow \left\{\begin{array}{l}2x^3-2x^2z+xz^2=2\\-2x^3+6x^2z-9xz^2+5z^3=14\\y=2z-x\end{array}\right.$

$\Leftrightarrow \left\{\begin{array}{l}2x^3-2x^2z+xz^2=2\\5z^3-16xz^2+20x^2z-16x^3=0\\y=2z-x\end{array}\right.$

As $x,z\ne0$ , put $t=\dfrac{z}{x}$ , we have:

$5t^3-16t^2+20t-16=0 \Leftrightarrow (t-2)(5t^2-6t+8)=0 \Leftrightarrow t=2$

Hence $z=2x$ and the system is equivalent to

$\left\{\begin{array}{l}2x^3-2x^2z+xz^2=2\\z=2x\\y=2z-x\end{array}\right.\Leftrightarrow \left\{\begin{array}{l}x=1\\y=3\\z=2\end{array}\right.$

So, the answer is $1+2+3=\boxed{6}$ .