3. Given real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=4$. Then the maximum value of $(2 x-y)^{2}+(2 y-z)^{2}+(2 z-x)^{2}$ is ( ).
(A) 12
(B) 20
(C) 28
(D) 36
3.C.

From the problem, we have
$$
\begin{array}{l}
(2 x-y)^{2}+(2 y-z)^{2}+(2 z-x)^{2} \\
\approx 20-4(x y+y z+z x) \\
=28-2(x+y+z)^{2} \leqslant 28 .
\end{array}
$$

When $x+y+z=0$, the equality holds.
Thus, the maximum value sought is 28.