$gt\Rightarrow1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$

$\Rightarrow\frac{1}{a^2}+1=\frac{1}{a^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{c}\right)$

$\frac{1}{ab}\sqrt{\frac{\left(a^2+1\right)\left(b^2+1\right)}{c^2+1}}=\sqrt{\frac{\left(1+\frac{1}{a^2}\right)\left(1+\frac{1}{b^2}\right)}{c^2\left(1+\frac{1}{c^2}\right)}}$

$=\frac{1}{c}.\sqrt{\frac{\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{a}\right)\left(\frac{1}{b}+\frac{1}{c}\right)}{\left(\frac{1}{c}+\frac{1}{a}\right)\left(\frac{1}{c}+\frac{1}{b}\right)}}=\frac{1}{c}\sqrt{\left(\frac{1}{a}+\frac{1}{b}\right)^2}$

$=\frac{1}{c}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{bc}+\frac{1}{ca}$

Tương tự với các cụm còn lại, ta được

$A=2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2$

$gt\Rightarrow1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$

$\Rightarrow\frac{1}{a^2}+1=\frac{1}{a^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{c}\right)$

$\frac{1}{ab}\sqrt{\frac{\left(a^2+1\right)\left(b^2+1\right)}{c^2+1}}=\sqrt{\frac{\left(1+\frac{1}{a^2}\right)\left(1+\frac{1}{b^2}\right)}{c^2\left(1+\frac{1}{c^2}\right)}}$

$=\frac{1}{c}.\sqrt{\frac{\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{a}\right)\left(\frac{1}{b}+\frac{1}{c}\right)}{\left(\frac{1}{c}+\frac{1}{a}\right)\left(\frac{1}{c}+\frac{1}{b}\right)}}=\frac{1}{c}\sqrt{\left(\frac{1}{a}+\frac{1}{b}\right)^2}$

$=\frac{1}{c}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{bc}+\frac{1}{ca}$

Tương tự với các cụm còn lại, ta được

$A=2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2$

Ta có:3¹⁰⁰⁰=9⁵⁰⁰ có không quá 500 chữ số .Kí hiệu tổng các chữ số của n là S(n),ta có:

A=S(3¹⁰⁰⁰)$\le$9.500=4500;B=S(A)<4+9+9+9=31.Vì 3¹⁰⁰⁰ chia hết cho 9 nên A,B,C đều chia hết cho 9

$\Rightarrow B\in\left\{9,18,27\right\}$ đều có tổng các chữ số là 9 nên C=9