For any two points $$M$$ and $$N$$ in the $$XY$$-plane, let $$\overrightarrow{MN}$$ denote the vector from $$M$$ to $$N$$, and $$\vec{0}$$ 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 $$\triangle PQR$$ such that

$$\overrightarrow{SP} + 5\,\overrightarrow{SQ} + 6\,\overrightarrow{SR} = \vec{0}.$$

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 ________.

Let the position vectors of the vertices be $$\mathbf{p},\;\mathbf{q},\;\mathbf{r}$$ for the points $$P,\;Q,\;R$$ respectively, and let $$\mathbf{s}$$ be the position vector of the point $$S$$.

The given vector condition is
$$\overrightarrow{SP}+5\,\overrightarrow{SQ}+6\,\overrightarrow{SR}=\vec{0}.$$

Using $$\overrightarrow{SP}=\mathbf{p}-\mathbf{s},\;\overrightarrow{SQ}=\mathbf{q}-\mathbf{s},\;\overrightarrow{SR}=\mathbf{r}-\mathbf{s},$$ we get

$$\bigl(\mathbf{p}-\mathbf{s}\bigr)+5\bigl(\mathbf{q}-\mathbf{s}\bigr)+6\bigl(\mathbf{r}-\mathbf{s}\bigr)=\vec{0}.$$

Simplifying,
$$\mathbf{p}+5\mathbf{q}+6\mathbf{r}-12\mathbf{s}= \vec{0}\;\Longrightarrow\; \mathbf{s}= \dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}}{12}.$$

Next, let $$E$$ and $$F$$ be the mid-points of $$PR$$ and $$QR$$.

Mid-point position vectors:
$$\mathbf{e}= \dfrac{\mathbf{p}+\mathbf{r}}{2}, \qquad \mathbf{f}= \dfrac{\mathbf{q}+\mathbf{r}}{2}.$$

Length of $$EF$$
$$\overrightarrow{EF}= \mathbf{f}-\mathbf{e}= \dfrac{\mathbf{q}+\mathbf{r}}{2}-\dfrac{\mathbf{p}+\mathbf{r}}{2} =\dfrac{\mathbf{q}-\mathbf{p}}{2}.$$
Hence $$|EF|=\dfrac{1}{2}\,|\mathbf{q}-\mathbf{p}|.$$

Length of $$ES$$
$$\overrightarrow{ES}= \mathbf{s}-\mathbf{e} =\dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}}{12}-\dfrac{\mathbf{p}+\mathbf{r}}{2} =\dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}-6\mathbf{p}-6\mathbf{r}}{12} =\dfrac{-5\mathbf{p}+5\mathbf{q}}{12} =\dfrac{5}{12}\,(\mathbf{q}-\mathbf{p}).$$
Therefore $$|ES|=\dfrac{5}{12}\,|\mathbf{q}-\mathbf{p}|.$$

Required ratio
$$\dfrac{|EF|}{|ES|}= \dfrac{\dfrac{1}{2}\,|\mathbf{q}-\mathbf{p}|}{\dfrac{5}{12}\,|\mathbf{q}-\mathbf{p}|} =\dfrac{1}{2}\times\dfrac{12}{5} =\dfrac{6}{5}=1.2.$$

Hence, $$\dfrac{\text{length of }EF}{\text{length of }ES}=1.2,$$ which lies in the range 1.15-1.25.