Guess, before proving

First, it is clear that $a, b, c > 1$

Multiplying both sides by $abc$ and factorise, we obtain:

$ab + bc + ca = abc$

$ab + ca - abc = -bc$

$a(b+c-ab) = -bc$

$a(ab -b-c) = bc$

$a((a-1)(b-1)-1) = bc$

$(a-1)(b-1) - 1 = \dfrac{bc}{a}$

Finally, we obtain $(a-1)(b-1) = \dfrac{bc}{a} + 1$

The LHS implies that $(a-1)(b-1)$ is a positive integer since $a-1$ and $b-1$ both are.

However, this is a contradiction from the RHS since $\dfrac{bc}{a}$ cannot be an integer, because $gcd(a,b) = gcd(a,c)=1$ .

Hence, there are $\boxed{0}$ sets of { $a,b,c$ } which satisfy the given equation.

1) If a = b = c = 3 the gcd of any pair is 3 so there are no solutions here.

2) If any pair are equal then we a gcd of more than one ( out of interest the solution would otherwise be 4,4,2)

3) Now Assume without loss of generality that $\ a > b > c \implies \frac{1}{a} < \frac{1}{b} < \frac{1}{c}$ . Now this means that: $\ 1 > \frac{1}{c} > \frac{1}{3} \implies 3 > c > 1 \therefore\ c = 2$ This leaves: $\ \frac{1}{a} + \frac{1}{b} = \frac{1}{2} \implies \frac{1}{2} > \frac{1}{b} > \frac{1}{4} \therefore\ b = 3 \ and \ a = 6$ However, gcd(a,b) = 3 so we have no solutions at all.