Let us keep dividing each of the numbers in the set $ \{a,b,c,n\}$ by $ 3$ until one of the numbers $ a,b,c,n$ is not divisible by $ 3$. Because dividing each of the numbers $ a,b,c,n$ by $ 3$ still gives the same equation, we will just work with the equation $ 6(6a^{2}+3b^{2}+c^{2}) = 5n^{2}$ assuming one of $ a,b,c,n$ is not divisible by $ 3$. Because left side is divisible by $ 3$ then the right side is divisible by $ 3$ so $ 3|5n^{2}$ but as $ 3$ and $ 5$ are coprime get $ 3|n^{2}$ and as $ 3$ is prime, $ 3|n$ therefore $ n=3k$ and $ 9|n^{2}$ so the right side is divisible by $ 9$. But then the left side is divisble by $ 9$ and as $ 3|6$ but $ 9$ does not divide $ 6$ we must have $ 3|(6a^{2}+3b^{2}+c^{2})$ . Because $ 3|(6a^{2}+3b^{2})$ have $ 3|c^{2}$ so $ 3|c$ so $ c=3l$ and $ c^{2}=9l^{2}$. Then, $ 6(6a^{2}+3b^{2}+c^{2}) = 5n^{2}\Leftrightarrow 6(6a^{2}+3b^{2}+9l^{2})=5*9k^{2}$ So, dividing both sides by $ 9$ obtain: $ 4a^{2}+2b^{2}+6l^{2}=5k^{2}$. Look at the solution $ (a,b,l,k)$ with $ k$ being minimal. Now, look at the equation mod $ 16$ -- then $ a^{2},b^{2},l^{2},k^{2}$ can be $ 0,1,4,9$ mod $ 16$ -- the set of quadratic residues of $ 16$. Now, left side is even, so right side is even so $ k$ is even, hence $ 5k^{2}$ is $ 0$ or $ 4$ mod $ 16$. Because $ 4|(4a^{2})$ then $ a^{2}$ is also $ 0$ or $ 4$ mod $ 16$. So, get $ 2b^{2}+6l^{2}$ is also $ 0$ or $ 4$ mod $ 16$, but now $ 2b^{2}$ is $ 2, 0, 8$ mod $ 16$ while $ 6l^{2}$ is $ 6, 0, 8$ mod $ 16$. By checking all cases, get $ 2b^{2}+6l^{2}$ can never be $ 4$ mod $ 16$ so it is $ 0$ mod $ 16$, but the only case when that is possible is if $ 2b^{2}$ and $ 6l^{2}$ are both $ 0$ mod $ 16$ or are $ 8$ and $ 8$ mod $ 16$ but this is possible only when $ b$ and $ l$ are both divisible by $ 2$. But then, as $ 2b^{2}+6l^{2}$ is $ 0$ mod $ 16$ then $ 4a^{2}$ is $ 0$ mod $ 16$ so $ a$ is also divisible by $ 2$. So, each of $ a,b,l,k$ is divisible by $ 2$ so $ (\frac{a}{2},\frac{b}{2},\frac{l}{2},\frac{k}{2})$ is also a solution in integers but because $ k$ is minimal, we have$ \frac{k}{2}=k$ so $ k=0$ and then $ a,b,l$ are each $ 0$ so $ a,b,c,n$ must be $ 0$ QED.

The rhs is divisible by $ 3$, so 3 must divide n. So $ 5n^2 - 36a^2 - 18b^2$ is divisible by 9, so 3 must divide c. We can now divide out the factor 9 to get: $ 5m^2 = 4a^2 + 2b^2 + 6d^2$. Now take m, a, b, d to be the solution with the smallest m, and consider residues mod 16. Squares = 0, 1, 4, or 9 mod 16. Clearly m is even so $ 5m^2 = 0$ or 4 mod 16. Similarly, $ 4a^2 = 0 or 4 mod 16$. Hence $ 2b^2 + 6d^2 = 0, 4 or 12 mod 16$. But $ 2b^2 = 0, 2 or 8 mod 16$ and $ 6d^2 = 0, 6 or 8 mod 16$. Hence $ 2b^2 + 6d^2 = 0, 2, 6, 8, 10 or 14 mod 16$. So it must be 0. So b and d are both even. So a cannot be even, otherwise $ \frac{m}{2};\frac{a}{2};\frac{b}{2};\frac{c}{2}$ would be a solution with smaller $ \frac{m^2}{2}$. So we can divide out the factor 4 and get: $ 5k^2 = a^2 + 2e^2 + 6f^2$ with a odd. Hence k is also odd. So $ 5k^2 - a^2 = 4 or 12 mod 16$. But we have just seen that $ 2e^2 + 6 f^2 cannot be 4 or 12 mod 16$. So there are no solutions.

Good solution, but can you please write your latex a bit more readable?