If $$\frac{\sqrt{60} + \sqrt{95} + \sqrt{228} + 12}{\sqrt{5} + 2\sqrt{12} + \sqrt{19}} = \frac{\sqrt{a} + \sqrt{b}}{c}$$, where $$a, b$$ are square-free natural numbers and $$c$$ is a natural number, then the value of $$a + b + 7c =$$

Put $$p = \sqrt{5}$$, $$q = \sqrt{12}$$, $$r = \sqrt{19}$$, so the numerator is $$pq + pr + qr + q^2 = (p + q)(q + r)$$ and the denominator is $$(p + q) + (q + r)$$. The quotient is therefore $$\frac{1}{\frac{1}{p + q} + \frac{1}{q + r}}$$, and rationalising gives $$\frac{1}{p+q} + \frac{1}{q+r} = \frac{2\sqrt{3} - \sqrt{5}}{7} + \frac{\sqrt{19} - 2\sqrt{3}}{7} = \frac{\sqrt{19} - \sqrt{5}}{7}$$. Hence the expression equals $$\frac{7}{\sqrt{19} - \sqrt{5}} = \frac{\sqrt{19} + \sqrt{5}}{2}$$, so $$a = 19$$, $$b = 5$$, $$c = 2$$ and $$a + b + 7c = 38$$.