Question 138
Find the value of the expression
$$(a+b+c) \left( \frac{1}{a+b-5c} + \frac{1}{b+c-5a} + \frac{1}{c+a-5b} \right),$$
if real $a$, $b$, $c$ satisfy the equality $$(a+b+c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = \frac{27}{2}$$ (all denominators are supposed to be different from zero).
$$(a+b+c) \left( \frac{1}{a+b-5c} + \frac{1}{b+c-5a} + \frac{1}{c+a-5b} \right),$$
if real $a$, $b$, $c$ satisfy the equality $$(a+b+c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = \frac{27}{2}$$ (all denominators are supposed to be different from zero).