Playing with $a,b$ and $c$

If $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}$ , then the left side is simplified to:

$\frac{ab+ac+bc}{abc} = \frac{1}{a+b+c}$

Cross-multiplying gives:

$(a+b+c)(ab+ac+bc) = abc$

Now:

$\begin{aligned} (a+b)(a+c)(b+c) &= a^{2}b + a^{2}c + ab^{2} + ac^{2} + b^{2}c + bc^{2} + 2abc \\ &= (a^{2}b + a^{2}c + ab^{2} + ac^{2} + b^{2}c + bc^{2} + 3abc) - abc \\ &= (a+b+c)(ab+ac+bc) - abc \end{aligned}$

Since $(a+b+c)(ab+ac+bc) = abc$ , then:

$(a+b)(a+c)(b+c) = 0$

as per the answer.