Find $x,y,z$ for the given conditions $$4x^2+25y^2+9z^2-10xy-15yz-6zx=0$$
$$x+y+z=5$$
I tried two approaches 
1) Substituting $x$ as $5-y-z$ in the first equation but didn't work out, I was getting $39y^2+19z^2-31yz-90y-70z+100=0$ which can't be factorized
2) First equation corresponds to $a^2+b^2+c^2-ab-bc-ca=0$, which means $a^3+b^3+c^3=3abc$, but didn't get a breakthrough.
I am stuck here, please help me.

Continuing your second method. 
Denote: $a=2x,b=5y,c=3z$, then:
$$\begin{cases}4x^2+25y^2+9z^2-10xy-15yz-6zx=0\\
x+y+z=5\end{cases} \Rightarrow \\
\begin{cases}a^2+b^2+c^2-ab-bc-ca=0\\
\frac a2+\frac b5+\frac c3=5\end{cases} \Rightarrow \\
2a^2+2b^2+2c^2-2ab-2bc-2ca=0 \Rightarrow \\
(a-b)^2+(b-c)^2+(c-a)^2=0 \Rightarrow \\
a=b=c=\frac{150}{31} \Rightarrow \\
(x,y,z)=\left(\frac{75}{31},\frac{30}{31},\frac{50}{31}\right)$$
WA answer.
