3. Let real numbers $x, y, z$ satisfy
$$
x^{2}+y^{2}+z^{2}-x y-y z-z x=27 \text {. }
$$

Then the maximum value of $|y-z|$ is $\qquad$
3. 6 .

The original equation is equivalent to a quadratic equation in $x$
$$
\begin{array}{l}
x^{2}-(y+z) x+y^{2}+z^{2}-y z-27=0 . \\
\text { And } \Delta=(y+z)^{2}-4\left(y^{2}+z^{2}-y z-27\right) \geqslant 0 \\
\Rightarrow(y-z)^{2} \leqslant 36 \Rightarrow|y-z| \leqslant 6 .
\end{array}
$$

When $|y-z|=6$, and $x=\frac{y+z}{2}$, the equality holds.
Therefore, the maximum value of $|y-z|$ is 6 .