1. Find all real numbers $p$ for which the system of inequalities

$$
\begin{aligned}
& 25+2 x^{2} \leqq 13 y+10 z-p \\
& 25+3 y^{2} \leqq 6 z+10 x \\
& 25+4 z^{2} \leqq 6 x+5 y+p
\end{aligned}
$$

with unknowns $x, y, z$ has a solution in the set of real numbers.
1. By adding all three inequalities, we obtain the inequality

$$
75+2 x^{2}+3 y^{2}+4 z^{2} \leqq 16 x+18 y+16 z
$$

from which, after "completing the square," we get

$$
2(x-4)^{2}+3(y-3)^{2}+4(z-2)^{2} \leqq 0 .
$$

This inequality, which is a consequence of the given system of inequalities, clearly holds only when the bases of all three squared terms on the left side of the inequality are zero, i.e., when $x=4$, $y=3$, $z=2$. Therefore, the given system has at most one solution for a given $p$, and that is the triplet of numbers written above. We will now determine for which value of the parameter $p$ this is indeed a solution. By substituting the values $x=4, y=3, z=2$ into the given system, we obtain the triplet of inequalities

$$
57 \leqq 59-p, \quad 52 \leqq 52, \quad 41 \leqq 39+p
$$

From the first inequality, we get the condition $p \leqq 2$, and from the third, the condition $p \geqq 2$. The number $p=2$ is therefore the only value of $p$ for which the given system has a solution in the set of real numbers.

For a complete solution, 6 points are awarded, 4 points for determining the unknowns $x, y, z$ and 2 points for determining the value of $p$.