1. The equation
$$
x^{4}+y^{4}+z^{4}-2 x^{2} y^{2}-2 y^{2} z^{2}-2 z^{2} x^{2}=24
$$

The set of all integer solutions $(x, y, z)$ is $\qquad$
$-1 . \varnothing$.
Notice,
$$
\begin{aligned}
x^{4}+ & y^{4}+z^{4}-2 x^{2} y^{2}-2 y^{2} z^{2}-2 z^{2} x^{2} \\
& =\left(x^{2}-y^{2}-z^{2}\right)^{2}-4 y^{2} z^{2} \\
& =\left(x^{2}-y^{2}-z^{2}+2 y z\right)\left(x^{2}-y^{2}-z^{2}-2 y z\right) \\
& =(x+y-z)(x-y+z)(x+y+z)(x-y-z) \\
& =24 .
\end{aligned}
$$

If the equation has integer solutions $(x, y, z)$, then
$$
x+y-z, x-y+z, x+y+z, x-y-z
$$

are all integers, and their product equals 24.
Since $x+y-z, x-y+z, x+y+z, x-y-z$ have the same parity, they are all even.
Therefore, the left side of the equation is divisible by 16, while the right side $16 \nmid 24$, leading to a contradiction.
Thus, the equation has no integer solutions.