阅读理解:在平面直角坐标系中,$P_{1}(x_{1}$,$y_{1})$,$P_{2}(x_{2}$,$y_{2})$,如何求$P_{1}P_{2}$的距离.如图,在$Rt\triangle P_{1}P_{2}Q$,$|P_{1}P_{2}|^{2}=|P_{1}Q|^{2}+|P_{2}Q|^{2}=(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$,所以$|P_{1}P_{2}|=\sqrt{{{({{x_2}-{x_1}})}^2}+{{({{y_2}-{y_1}})}^2}}$.因此,我们得到平面上两点$P_{1}(x_{1}$,$y_{1})$,$P_{2}(x_{2}$,$y_{2})$之间的距离公式为$|P_{1}P_{2}|=\sqrt{{{({{x_2}-{x_1}})}^2}+{{({{y_2}-{y_1}})}^2}}$.根据上面得到的公式,解决下列问题:
$(1)$已知点$P\left(2,6\right)$,$Q\left(-3,-6\right)$,试求$P$、$Q$两点间的距离；
$(2)$已知点$M\left(m,5\right)$,$N\left(1,2\right)$且$MN=5$,求$m$的值；
$(3)$求代数式$\sqrt{{{({x-3})}^2}+{y^2}}+\sqrt{{{({x+3})}^2}+{{({y+4})}^2}}$的最小值.