[Solution] Let the left arm length of the balance be $l_{1}$, and the right arm length be $l_{2}$; the weight of the left pan be $m_{1}$, and the right pan be $m_{2}$. According to the problem, we have
$$\left\{\begin{array}{l} \left(A+m_{1}\right) l_{1}=\left(A_{1}+m_{2}\right) l_{2}, \\ \left(B+m_{1}\right) l_{1}=\left(B_{1}+m_{2}\right) l_{2}, \\ \left(C+m_{1}\right) l_{1}=\left(C_{1}+m_{2}\right) l_{2}, \\ \left(A_{2}+m_{1}\right) l_{1}=\left(A+m_{2}\right) l_{2}, \\ \left(B_{2}+m_{1}\right) l_{1}=\left(B+m_{2}\right) l_{2} . \end{array}\right.$$

Let $m_{2} l_{2}-m_{1} l_{1}=k$, rearranging the above system of equations gives
$$\left\{\begin{array}{l} A l_{1}=A_{1} l_{2}+k, \\ B l_{1}=B_{1} l_{2}+k, \\ C l_{1}=C_{1} l_{2}+k \\ A_{2} l_{1}=A l_{2}+k, \\ B l_{1}=B l_{2}+k . \end{array}\right.$$

From (1) - (2) we get $(A-B) l_{1}=\left(A_{1}-B_{1}\right) l_{2}$,
From (4) - (5) we get $\left(A_{2}-B_{2}\right) l_{1}=(A-B) l_{2}$,
Therefore, (assuming $A>B$) we have $\frac{l_{1}}{l_{2}}=\sqrt{\frac{A_{1}-B_{1}}{A_{2}-B_{2}}}$.
From (1) and (4) we get $A=\frac{A_{1} l_{2}+A_{2} l_{1}}{l_{1}+l_{2}}=\frac{A_{1}+A_{2} \frac{l_{1}}{l_{2}}}{\frac{l_{1}}{l_{2}}+1}$.
Substituting (6) into the above equation and simplifying, we get
$$A=\frac{A_{1} \sqrt{A_{2}-B_{2}}+A_{2} \sqrt{A_{1}-B_{1}}}{\sqrt{A_{1}-B_{1}}+\sqrt{A_{2}-B_{2}}} .$$

From (1) and (3) we get $C=\left(C_{1}-A_{1}\right) \cdot \frac{l_{2}}{l_{1}}+A$.
Substituting (6) and (7) into the above equation, we get
$$C=\left(C_{1}-A_{1}\right) \sqrt{\frac{A_{2}-B_{2}}{A_{1}-B_{1}}}+\frac{A_{1} \sqrt{A_{2}-B_{2}}+A_{2} \sqrt{A_{1}-B_{1}}}{\sqrt{A_{1}-B_{1}}+\sqrt{A_{2}-B_{2}}} .$$