Correct Answer: 1

Solution : $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}$
$⇒a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc}...(i)$
$⇒b-c=\frac{1}{a}-\frac{1}{c}=\frac{c-a}{ac}...(ii)$
$⇒a-c=\frac{1}{a}-\frac{1}{b}=\frac{b-a}{ba}...(iii)$
Multiplying the above equations, we have,
$⇒(a-b)(b-c)(a-c)=(\frac{b-c}{bc})(\frac{c-a}{ac})(\frac{b-a}{ba})$
$⇒(a-b)(b-c)(a-c)=(\frac{(b-c)(c-a)(b-a)}{a^2b^2c^2})$
$⇒ a^2b^2c^2=(\frac{(b-c)(c-a)(b-a)}{(a-b)(b-c)(a-c)})$
$a \neq b\neq c\neq 0$
$⇒ a^2b^2c^2= 1$
Hence, the correct answer is 1.