If $2x^2+5y^2+2z^2-2xy-6yz=0$, then $x:y:z$ is:

Let's rewrite and rearrange the given equation to group the terms into perfect squares:
$$2x^2+5y^2+2z^2-2xy-6yz=0$$

Multiply the entire equation by $2$:

$$4x^2+10y^2+4z^2-4xy-12yz=0$$

Now split $10y^2$ into $y^2+9y^2$:

$$(4x^2-4xy+y^2)+(9y^2-12yz+4z^2)=0$$

Recognizing the perfect square identities $(a-b{)}^2=a^2-2ab+b^2$:

$$(2x-y{)}^2+(3y-2z{)}^2=0$$

Since the sum of two real squares equals zero, each term must independently equal zero:

1. $2x-y=0⟹2x=y⟹\frac{x}{y}=\frac{1}{2}$
2. $3y-2z=0⟹3y=2z⟹\frac{y}{z}=\frac{2}{3}$

To match the ratios, choose a common value for $y$. Here $y=2$ fits perfectly:

* If $y=2$, then $2x=2⟹x=1$
* If $y=2$, then $2z=3(2)⟹z=3$

Thus, the ratio $x:y:z$ is $1:2:3$.