Prove there isn't $r_1, r_2,r_3 \in \mathbb{Q}$ so that ${r_1}^2 + {r_2}^2 + {r_3}^2=7 \tag1$
From (1) we get $a^2 + b^2 + c^2=7n^2 \tag2$ where $a,b,c,n \in \mathbb{N}$. I have tried playing with parity of these numbers, without success.
UPDATE
Suppose $n$ is even. Then either $a, b, c$ are all even or only one of them, let's say $a$ is even. The latest case is not possible because applying modulo 4 to (2) we get $2=0$. So $a, b, c$ are all even. Repeatedly simplifying by 4, we can reduce this case to $n$ odd.
Now suppose $n$ is odd. Then either $a, b, c$ are all odd or two of them, let's say $a,b$ are even. The latest case is not possible because applying modulo 4 to (2) we get $1=3$.
The only case I cannot cover is $a,b,c,n$ all odd.
This is about 2-adic restrictions. First, odd squares of integers are $1 \pmod 8.$ Integer squares can only be $0,1,4 \pmod 8$ in any case. Therefore the sum of three integer squares cannot be $7 \pmod 8.$
Next, if the sum of three squares is divisible by $4,$ so $x^2 + y^2 + z^2 = k$ with $k \equiv 0 \pmod 4,$ then $x,y,z$ must be even so we can divide through and get integers $\left( \frac{x}{2} \right)^2 + \left( \frac{y}{2} \right)^2 +\left( \frac{z}{2} \right)^2 = \frac{k}{4}.$ This is all you need to deal with $x^2 + y^2 + z^2 = 7 n^2$ in integers.
Also worth mentioning Aubry-Davenport-Cassels, there is a geometric proof that, if a number is the sum of three rational squares, it is also the sum of three integer squares. This is presented in Serre's little book.
About $7$ itself, if we have $u^2 + v^2 + w^2 = k$ with $u,v,w$ not all divisible by $7,$ then we can solve $x^2 + y^2 + z^2 = 49k$ with $x,y,z$ all nonzero $\pmod 7.$ That is, we choose one of $(u,v,w)$ or $(-u,v,w)$ or $(u,-v,w)$ or $(u,v,-w)$ (and rename as $(u,v,w)$ again) so that $u + 2 v + 4 w \neq 0 \pmod 7.$ Then we take $$ x = 3u+6v - 2w, \; \; \; y = -2u+3v +6w, \; \; \; z = 6u -2v +3 w. $$ All are nonzero $\pmod 7.$ This is the rational orthogonal matrix $$ \frac{1}{7} \left( \begin{array}{rrr} 3 & 6 & -2 \\ -2 & 3 & 6 \\ 6 & -2 & 3 \end{array} \right) $$ as in PALL 1940