10. (15 points) Solve the system of equations
$$
\left\{\begin{array}{l}
a b+c+d=3 \\
b c+d+a=5 \\
c d+a+b=2 \\
d a+b+c=6
\end{array}\right.
$$
$$
\begin{array}{l}
\text { 10. (1) +(2)-(3) - (4) gives } \\
(a+c)(b-d)+2(d-b)=0 \\
\Rightarrow(b-d)(a+c-2)=0 .
\end{array}
$$

If $b-d=0$, then $b=d$, which contradicts equations (1) and (4).
If $a+c-2=0$, then
$$
a+c=2 \text {. }
$$

By adding (3) and (4) and combining with equation (5), we get
$$
b+d=3 \text {. }
$$

By adding (2) and (3) and combining with equation (6), we get
$$
2 a+3 c=4 \text {. }
$$

From equations (5) and (7), we have $a=2, c=0$.
Substituting into equation (2) gives $d=3$. Thus, $b=0$.
Therefore, $(a, b, c, d)=(2,0,0,3)$.