Set $a+b+c = x$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = y$. Let $k = \frac{x}{6}$, then
$$a+b-5c=6(k-c), \quad b+c-5a=6(k-a), \quad c+a-5b=6(k-b).$$
It follows that
$$(a+b+c) \left( \frac{1}{a+b-5c} + \frac{1}{b+c-5a} + \frac{1}{c+a-5b} \right) = \frac{x}{6} \left( \frac{1}{k-a} + \frac{1}{k-b} + \frac{1}{k-c} \right). \quad (1)$$
Adding the fractions in the right-hand side of (1), we obtain $\frac{A}{B}$, where
$$A = (k-a)(k-b) + (k-b)(k-c) + (k-c)(k-a) \text{ and } B = (k-a)(k-b)(k-c). \quad (2)$$
Let $z = abc$, then $ab + bc + ca = yz$. From (2) it follows
$$A = 3k^2 - 2zk + yz \quad \text{and}\quad B = k^3 - zk^2 + yz - z. \quad (3)$$
Since $x = 6k$ and $xy = \frac{27}{2}$, we have $y = \frac{9}{4k}$. Substitute for $x$ and $y$ in (3) to obtain
$$A = 3k^2 - 2 \cdot 6k \cdot k + \frac{9}{4k}z = 9 \left(-k^2 + \frac{z}{4k}\right), \quad B = k^3 - 6k \cdot k^2 + \frac{9}{4k}zk - z = 5k \left(-k^2 + \frac{z}{4k}\right).$$
Thus
$$\frac{A}{B} = \frac{9}{5k}, \quad \text{and}\quad \frac{x}{6} \cdot \frac{A}{B} = k \cdot \frac{9}{5k} = \frac{9}{5}.$$
Hence
$$(a+b+c) \left( \frac{1}{a+b-5c} + \frac{1}{b+c-5a} + \frac{1}{c+a-5b} \right) = \frac{9}{5}.$$