2. Real numbers $x, y, z$ satisfy the relations:

$$
4 x^{2}-2 x-30 y z=25 y^{2}+5 y+12 x z=9 z^{2}-3 z-20 x y .
$$

Find the maximum of the sum $a+b+c$, where $a=2 x+5 y, b=3 z+5 y, c=3 z-2 x$.
Solution. Note that

$$
a-b+c=0
$$

Let $A=4 x^{2}-2 x-30 y z, B=25 y^{2}+5 y+12 x z$ and $C=9 z^{2}-3 z-20 x y$. Subtracting these equations from each other, we get

$$
\begin{aligned}
& A-B=a \cdot(2 x-6 z-5 y-1)=0 \\
& B-C=b \cdot(5 y+4 x-3 z+1)=0 \\
& A-C=c \cdot(1-2 x-10 y-3 z)=0
\end{aligned}
$$

Assume that all three numbers $a, b, c$ are non-zero. Then $2 x-6 z-5 y-1=0, 5 y+4 x-3 z+1=0$ and $1-2 x-10 y-3 z=0$, which is impossible, as adding the second equation to the third and subtracting the first, we get $3=0$. Therefore, at least one of the numbers $a, b, c$ is zero. Consider the possible cases:

1) All three numbers $a, b, c$ are zero. The triplet $a=b=c=0$ clearly satisfies the conditions of the problem (it is sufficient to take $x=y=z=0$).
2) Only two of the numbers $a, b, c$ are zero. This is impossible: if two numbers are zero, then, according to (1), the third is also zero.
3) Only one of the numbers $a, b, c$ is zero:

- $a=0$. Then $x=-\frac{5 y}{2}$. From system (2) we find $b=c=1$;

## Interregional Olympiad for Schoolchildren Based on Departmental Educational Organizations

 in mathematics- $b=0$. Then $a=-c=1$;
- $c=0$. Then $a=b=-1$.

Thus, there are 4 possible options: $(0,0,0),(0,1,1),(1,0,-1),(-1,-1,0)$.

Answer: 2.