Let $k$ be a positive integer such that $$k=\frac{a_1^2+a_2^2 + a_3^2}{b_1+b_2+b_3}=\frac{d^2+(2d)^2+(3d)^2}{d^2 + d^2\cdot q + d^2\cdot q^2}=\frac{14}{1+q+q^2}$$ This means the following equation has a rational root $0 < q < 1$: \begin{equation} k\cdot q^2 + k\cdot q +(k-14)=0 \end{equation} Firstly, because \myJustRef{eq_22} is solvable in rational numbers, its discriminant must be non-negative. $$k^2 - 4\times k \times (k-14)\ge 0 \implies k = 1, 2, 3, 4$$ Next, suppose $q=\frac{m}{n}$ where $m < n$ are two positive integers. Then $n\ge 2$. By the Rational Zero Theorem on page \pageref{theorem_rational_zero}, it must hold that $$n\mid k \quad \text{and}\quad m\mid (k-14)$$ Do casework. 1) when $k=1$, no solution. 2) when $k=2$, $n=2$, $m=1$. Thus $q={\frac{1}{2}}$. 3) when $k=3$, $n=3$, $m=1$. Thus $q={\frac{1}{3}}$. 4) when $k=4$, $n=4$, $m=2$. Thus $q={2}{4}={\frac{1}{2}}$ Setting $q$ as $\frac{1}{2}$ and $\frac{1}{3}$ finds $\boxed{\frac{1}{2}}$ is the only solution.