Solution:
It can be easily seen that
$$9 n = (2 b + 2 c - a)^{2} + (2 c + 2 a - b)^{2} + (2 a + 2 b - c)^{2}$$
Thus we can take $p_{1} = p_{2} = p_{3} = 2$, $q_{1} = q_{2} = q_{3} = 2$ and $r_{1} = r_{2} = r_{3} = -1$. Suppose $3$ does not divide $\gcd(a, b, c)$. Then $3$ does divide at least one of $a, b, c$; say $3$ does not divide $a$. Note that each of $2 b + 2 c - a$, $2 c + 2 a - b$ and $2 a + 2 b - c$ is either divisible by $3$ or none of them is divisible by $3$, as the difference of any two sums is always divisible by $3$. If $3$ does not divide $2 b + 2 c - a$, then we have the required representation. If $3$ divides $2 b + 2 c - a$, then $3$ does not divide $2 b + 2 c + a$. On the other hand, we also note that
$$9 n = (2 b + 2 c + a)^{2} + (2 c - 2 a - b)^{2} + (-2 a + 2 b - c)^{2} = x^{2} + y^{2} + z^{2}$$
where $x = 2 b + 2 c + a$, $y = 2 c - 2 a - b$ and $z = -2 a + 2 b - c$. Since $x - y = 3(b + a)$ and $3$ does not divide $x$, it follows that $3$ does not divide $y$ as well. Similarly, we conclude that $3$ does not divide $z$.