This kind of systems can be tipically done with multipliyng by "convenient ones":
$$\begin{array}{rl}
(1)&8x=\lambda (y+3z)\,\,\,\,/\cdot x\\
(2)&6y=\lambda (x+2z)\,\,\,\,/\cdot y\\
(3)&10z=\lambda (2y+3x)\,\,\,\,/\cdot z\\
(4)&xy+2yz+3xz=6\\
\end{array}$$
I've selected these "convenient ones" because they'll lead into terms with $xy$, $xz$ and $yz$ on the RHS of equations $(1)$, $(2)$ and $(3)$.
Then
$$\begin{array}{rl}
(1)\Rightarrow& \frac{8x^2}{\lambda}=xy+3xz\,\,\,(5)\\
(2)\Rightarrow& \frac{6y^2}{\lambda}=xy+2yz\,\,\,(6)\\
(3)\Rightarrow& \frac{10z^2}{\lambda}=2yz+3xz\,\,\,(7)\\
\end{array}$$
Summing $(5)+(6)+(7)$,
$$\begin{array}{rcl}
\frac{8x^2}{\lambda}+\frac{6y^2}{\lambda}+\frac{10z^2}{\lambda}&=&2xy+4yz+6xz\\
\Rightarrow\frac{1}{\lambda}(4x^2+3y^2+5y^2)&=&(xy+2yz+3xz)\,\,\,(8)
\end{array}$$
Replacing $(4)$ in $(8)$,
$$(4x^2+3y^2+5y^2)=6\lambda\,\,\,(9)$$
EDIT:
This yields an expression that we will use later.
Now, let's replace $(1)$, $(2)$ and $(3)$ in $(4)$, simplifying their LHS for $x$, $y$ and $z$, respectively:
$$\frac{\lambda(y+3z)}{8} \frac{\lambda(x+2z)}{6}+2 \frac{\lambda(x+2z)}{6} \frac{\lambda(2y+3x)}{10} +3\frac{\lambda(y+3z)}{8}\frac{\lambda(2y+3x)}{10}=6$$
After a bit of simplifying (and using $(4)$ a couple of times and $(9)$ once) you'll get a cubic equation for $\lambda$:
$$\lambda^2(\lambda+8)=40$$
The only real root for this equation is $\lambda=2$.
Now you can find $x$, $y$ and $z$.
Good luck!!
