3. Given that the four sides $a, b, c, d$ of quadrilateral $A B C D$ satisfy
$$
\left\{\begin{array}{l}
a-3 b+2 c=0, \\
b-3 c+2 d=0, \\
c-3 d+2 a=0, \\
d-3 a+2 b=0 .
\end{array}\right.
$$

then the quadrilateral $A B C D$ is ().
(A) rhombus
(B) cyclic quadrilateral
(C) rectangle
(D) trapezoid
3. D.

In the equation, we know
$$
\begin{array}{l}
a-b=2(b-c), \\
b-c=2(c \cdots d), \\
c-a=2(d-a) . \\
d-a=2(a-b) .
\end{array}
$$

From (1) and (3), substituting (2) into (1) yields
$$
\begin{aligned}
a \cdots b & =2(b-c)=2^{2}(c-d)=2^{3}(d-a) \\
& =2^{4}(a-b) .
\end{aligned}
$$

Thus, $a-b=0$, which means $a=b$. Combining (1) and (2), we conclude $a=b=c=d$.