For any two points M and N in the XY-plane, let $\overline{MN}$ denote the vector from M to N, and $\overline{0}$ denote the zero vector. Let P, Q and R be three distinct points in the XY-plane. Let S be a point inside the triangle $\Delta PQR$ such that $\overline{SP} + 5\overline{SQ} + 6\overline{SR} = \overline{0}$. Let E and F be the mid-points of the sides PR and QR, respectively. Then the value of $\frac{\text{length of the line segment } EF}{\text{length of the line segment } ES}$ is ______.

$\text{Given: } P,\ Q,\ R \text{ distinct points in the plane}$

S inside triangle PQR: S̅P̅ + 5 S̅Q̅ + 6 S̅R̅ = 0̅

$E,\ F \text{ mid-points of } PR \text{ and } QR$

$\text{To find: } \dfrac{EF}{ES}$

$1.\ \text{midpoint theorem links } EF \text{ and } PQ$

$2.\ \text{take } S \text{ as origin, use } \overline{SP}+5\overline{SQ}+6\overline{SR}=\overline{0}$

$3.\ \text{compute } ES, \text{ then the ratio}$

$E,\ F \text{ midpoints of } PR,\ QR$

$\Rightarrow\ EF \parallel PQ,\qquad EF = \dfrac{1}{2}\,PQ$

$\text{Origin at } S:\quad \vec{SP}=a,\quad \vec{SQ}=b,\quad \vec{SR}=c$

$a + 5b + 6c = 0$

$a = -5b - 6c$

$\vec{SE} = \dfrac{a+c}{2}$

$\vec{SF} = \dfrac{b+c}{2}$

$b - a = b - (-5b-6c) = 6b + 6c = 6(b+c)$

$a + c = -5b - 6c + c = -5(b+c)$

$ES = |\vec{SE}| = \dfrac{|a+c|}{2} = \dfrac{5}{2}\,|b+c|$

$EF = \dfrac{|b-a|}{2} = \dfrac{6\,|b+c|}{2} = 3\,|b+c|$

$\dfrac{EF}{ES} = \dfrac{3\,|b+c|}{\tfrac{5}{2}\,|b+c|} = \dfrac{6}{5}$

$\boxed{\dfrac{EF}{ES} = \dfrac{6}{5}}$