Show that $6(6a^{2}+3b^{2}+c^{2})=5n^{2}$ has No Solution in Integers Except for the Trivial Solution

Question

Prove that the equation $$6(6a^2+3b^2+c^2)=5n^2$$ has no solution in integers except $$a=b=c=n=0$$ Now as we can see that $6|5n^2=>6|n$ So let $n=6n_1$ This gives $$6(6a^2+3b^2+c^2)=5*36{n_1}^2$$ $$=>6a^2+3b^2+c^2=30{n_1}^2$$ This implies $3|c$ So let $c=3c_1$ This gives $$6a^2+3b^2+9{c_1}^2=30{n_1}^2$$ $$=>2a^2+b^2+3{c_1}^2=10{n_1}^2-(1)$$ This gives $2|b^2+3{c_1}^2$ So either $b$ and $c_1$ are both odd or both even Case 1: When $b$ and $c_1$ are both odd then if $a$ is even then looking modulo $8$ our lhs is $2a^2+b^2+3{c_1}^2$ is congruent to $0+1+3=4$ mod$8$ But $10{n_1}^2$ is congruent to $0, 2$ mod$8$ So no solution with even $a$ Now for odd $a$ we get our lhs $2a^2+b^2+3{c_1}^2$ is congruent to $2+1+3=6$ mod$8$ But again $10{n_1}^2$ is congruent to $0,2$ mod$8$ So again no solution Case 2: When $b$ and $c_1$ are both even then let $b=2b_1$ and $c_1=2c_2$ then our equation becomes $$2a^2+4{b_1}^2+12{c_2}^2=10{n_1}^2$$ $$=>a^2+2{b_1}^2+6{c_2}^2=5{n_1}^2$$ Now $2|5{n_1}^2-a^2$ So either $a$ and $n_1$ are both even or both odd So again Case 1: When both $a$ and $n_1$ are both odd then if $b_1$ is odd then modulo $8$ Now we see that if $b_1$ is odd then $5{n_1}^2-a^2-2{b_1}^2$ is congruent to $5-1-2=2$ mod$8$ but $5{n_1}^2-a^2-2{b_1}^2=6{c_2}^2$ is congruent to $0, 6$ mod$8$ but this is not possible and if $b_1$ is even then again $5{n_1}^2-a^2-2{b_1}^2$ is congruent to $5-1-0=4$ mod$8$ but again this is not possible so Case 2: When both $a$ and $n_1$ are both even then let $a=2a_1$ and $n_1=2n_2$ then our equation becomes $$4{a_1}^2+2{b_1}^2+6{c_2}^2=20{n_2}^2$$ $$=>2{a_1}^2+{b_1}^2+3{c_2}^2=10{n_2}^2$$ This is similar to equation $(1)$ Now if we repeat this process then we again get a infinite set of decreasing positive integers satisfying the equation which is not possible by Fermat's Infinite Descent Principal. So there are no positive integer solutions to the equation. As it is easy to see that whenever $(-a, -b, -c, -d)$ is a solution to the given equation where $a, b, c, d$ are natural numbers then $(a, b, c, d)$ is also a solution but as the equation has no solution in positive integers so we can conclude that the given equation has no solution in integers. Is The Proof Correct?