[proofplan] Assume a rational point exists and rescale it to a primitive integer solution. Reducing the equation modulo $7$ forces both integer coordinates $x$ and $y$ to be divisible by $7$, because the only quadratic residues modulo $7$ are $0,1,2,4$ and no nonzero square is the negative of another square modulo $7$. Substituting this divisibility back into the original equation forces $z$ to be divisible by $7$ as well, contradicting primitivity. [/proofplan]

[step:Clear denominators to obtain a primitive integer solution] Assume, for contradiction, that $C(\mathbb{Q})$ is nonempty. Then there exist rational numbers $X,Y,Z \in \mathbb{Q}$, not all zero, satisfying \begin{align*} X^2 + Y^2 = 7Z^2. \end{align*} Choose a positive integer $d \in \mathbb{N}$ such that $dX,dY,dZ \in \mathbb{Z}$. Define \begin{align*} a := dX, \qquad b := dY, \qquad c := dZ. \end{align*} Then $(a,b,c) \in \mathbb{Z}^3$ is not the zero triple and satisfies \begin{align*} a^2 + b^2 = 7c^2. \end{align*} Let $g := \gcd(a,b,c)$, and define \begin{align*} x := \frac{a}{g}, \qquad y := \frac{b}{g}, \qquad z := \frac{c}{g}. \end{align*} Then $(x,y,z) \in \mathbb{Z}^3$ is not the zero triple, $\gcd(x,y,z)=1$, and division of the equation $a^2+b^2=7c^2$ by $g^2$ gives \begin{align*} x^2 + y^2 = 7z^2. \end{align*} [/step]

[step:Reduce the primitive equation modulo $7$] Reduce \begin{align*} x^2 + y^2 = 7z^2 \end{align*} modulo $7$. Since $7z^2 \equiv 0 \pmod{7}$, we obtain \begin{align*} x^2 + y^2 \equiv 0 \pmod{7}. \end{align*} The quadratic residues modulo $7$ are computed from the residue classes $0,1,2,3,4,5,6$: \begin{align*} 0^2 &\equiv 0, & 1^2 &\equiv 1, & 2^2 &\equiv 4, & 3^2 &\equiv 2, \\ 4^2 &\equiv 2, & 5^2 &\equiv 4, & 6^2 &\equiv 1 \pmod{7}. \end{align*} Thus every square modulo $7$ belongs to $\{0,1,2,4\}$. [/step]

[step:Use the residue computation to force $7 \mid x$ and $7 \mid y$] From \begin{align*} x^2 + y^2 \equiv 0 \pmod{7}, \end{align*} the residue class of $x^2$ must be the additive inverse of the residue class of $y^2$ modulo $7$. Among the quadratic residues $\{0,1,2,4\}$, the only pair whose sum is congruent to $0$ modulo $7$ is $(0,0)$, because \begin{align*} 1+1 &\equiv 2, & 1+2 &\equiv 3, & 1+4 &\equiv 5, \\ 2+2 &\equiv 4, & 2+4 &\equiv 6, & 4+4 &\equiv 1 \pmod{7}. \end{align*} Therefore \begin{align*} x^2 \equiv 0 \pmod{7} \qquad\text{and}\qquad y^2 \equiv 0 \pmod{7}. \end{align*} By the same residue computation, a square is congruent to $0$ modulo $7$ exactly when its base is congruent to $0$ modulo $7$. Hence \begin{align*} 7 \mid x \qquad\text{and}\qquad 7 \mid y. \end{align*} [/step]

[step:Substitute the divisibility back into the equation and contradict primitivity] Since $7 \mid x$ and $7 \mid y$, choose integers $r,s \in \mathbb{Z}$ such that \begin{align*} x = 7r, \qquad y = 7s. \end{align*} Substituting into $x^2+y^2=7z^2$ gives \begin{align*} 49r^2 + 49s^2 = 7z^2. \end{align*} Dividing by $7$ yields \begin{align*} z^2 = 7(r^2+s^2), \end{align*} so $7 \mid z^2$. Again using the residue computation modulo $7$, this implies $7 \mid z$. We have shown that $7$ divides each of $x,y,z$, contradicting $\gcd(x,y,z)=1$. Therefore no primitive integer solution exists, and hence no rational point exists on $C$. This proves \begin{align*} C(\mathbb{Q}) = \varnothing. \end{align*} [/step]