Let the $x$ coordinates of these points are $x_a$, $x_c$, $x_e$, $x_b$, $x_d$, and $x_f$, respectively. Then, it must have $$x_a < x_b < x_d < x_c < x_e < x_f$$ and $$P(x)=(x-x_a)(x-x_c)(x-x_e)$$ and $$P(x)-m=(x-x_b)(x-x_d)(x-x_f)$$ Meanwhile, $$ \left\{ \begin{array}{rlcrl} (x_a-x_b)^2 + m^2 &= 7 &\implies & x_a - x_b &=\sqrt{7-m^2}\\ (x_c-x_d)^2 + m^2 &=15&\implies & x_c - x_d &=-\sqrt{15-m^2}\quad\scriptsize{(\because x_c < x_d)}\\ (x_e-x_f)^2 + m^2 &=10 &\implies & x_e-x_f & =\sqrt{10-m^2} \end{array} \right. $$ Adding these three equations gives $$(x_a - x_b)+(x_c-x_d)+(x_e-x_f) =\sqrt{7-m^2}-\sqrt{15-m^2}+\sqrt{10-m^2}$$ Applying Vieta's theorem on both $P(x)$ and $P(x)-m$ gives $x_a+x_c+x_e=x_b+x_d+x_f$ because both equals to the coefficient of the term $x^2$. Setting this result to the previous equation yields $$\sqrt{7-m^2}-\sqrt{15-m^2}+\sqrt{10-m^2}=0\implies m=\boxed{\sqrt{6}}$$