If (x + y + z) = 0, then what is the value of 3y^2 + x^2 + z^22y^2 - xz?

Answer: Option A

$$\begin{array}{rl}& x+y+z=0,\frac{3y^2+x^2+z^2}{2y^2-xz}=?\\ & \text{Put }x=1,y=1,z=-2\\ & \Rightarrow \frac{3{\left(y\right)}^2+x^2+z^2}{2y^2-xz}\\ & \Rightarrow \frac{3{\left(1\right)}^2+{\left(1\right)}^2+{\left(-2\right)}^2}{2{\left(1\right)}^2-\left(1\times \left(-2\right)\right)}\\ & \Rightarrow \frac{4+4}{2+2}\\ & \Rightarrow \frac{8}{4}\\ & \Rightarrow 2\end{array}$$