$x_1$ and $x_2$, $y_1$ and $y_2$, $z_1$ and $z_2$ are distinct to one another and respectively roots of $m_1x^2 + n_1x + p_1 = 0$, $m_2y^2 + n_2y + p_2 = 0$, $m_3z^2 + n_3z + p_3 = 0$. Prove that $x_1 + x_2 = y_1 + y_2 = z_1 + z_2$ if the following condition is achieved
$$\begin{align} &(m_2 + m_3)(x_1^2 - x_2^2) + (n_2 + n_3)(x_1 - x_2) + (p_2 + p_3)\\ = &(m_3 + m_1)(y_1^2 - y_2^2) + (n_3 + n_1)(y_1 - y_2) + (p_3 + p_1)\\ = &(m_1 + m_2)(z_1^2 - z_2^2) + (n_1 + n_2)(z_1 - z_2) + (p_1 + p_2) = 0\\ \end{align}$$
Of course, this problem is adapted from a recent competition. And I can't solve it indeed. But using the Vièta formulae, I hope that I can solve this, but I just don't know how.
Let $P_i(X)=m_iX^2+n_iX+p_i$, $i=1,2,3$. Then $$ P_i(x_1)-P_i(x_2)=(x_1-x_2)(m_i(x_1+x_2)+n_i). $$ But $x_1+x_2=-n_1/m_1$, so $$ P_i(x_1)-P_i(x_2)=\frac{x_1-x_2}{m_1}\cdot(n_im_1-m_in_1). $$ Summing over all $i$, $$ \sum_i [P_i(x_1)-P_i(x_2)]=\frac{x_1-x_2}{m_1}\cdot\sum_i(n_im_1-m_in_1). $$
So the given condition $$ (m_2+m_3)(x_1^2-x_2^2)+(n_2+n_3)(x_1-x_2)+(p_2+p_3)=\dots=0 $$ is equivalent to $$ \sum_i(n_im_j-m_in_j)=0\quad\forall j $$ But that is $$ \frac{n_j}{m_j}=\frac{\sum_{i\neq j} n_i}{\sum_{i\neq j} m_i}=\frac{\sum_i n_i}{\sum_i m_i} $$ so all $n_j/m_j$ are equal, i.e., $x_1+x_2=y_1+y_2=z_1+z_2$, as claimed.