Turns out the details are easy so I worked them out myself :) The highlighted statement is true.
Let me assume $4\mid M$. Pick $p_2,p_3$ arbitrary satisfying the congruence modulo $M$ (they exist by Dirichlet). Take any $p_1$ which is congruent to $r_1\pmod M$, congruent to $p_2^{-1}\pmod{p_3}$, and such that $$\left(\frac{p_1}{p_2}\right)=\varepsilon\cdot(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}}$$ (which exists by Dirichlet, CRT, and existence of (non)residues modulo $p_2$.) We have $$\begin{align*}\left(\frac{p_2}{p_3}\right)\left(\frac{p_3}{p_1}\right)&=\left(\frac{p_2}{p_3}\right)\left(\frac{p_1}{p_3}\right)\cdot(-1)^{\frac{p_1-1}{2}\frac{p_3-1}{2}}\\ &=\left(\frac{p_1p_2}{p_3}\right)\cdot(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}}\\ &=\left(\frac{1}{p_3}\right)\cdot(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}}\\ &=(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}} \end{align*}$$ (second inequality follows since $p_1\equiv r_1,p_3\equiv r_3\pmod 4$), so multiplying by $\left(\frac{p_1}{p_2}\right)$ leaves $\varepsilon$.
