Step-by-Step Solution

1. Represent the points using position vectors. Let the origin be denoted by $O$. The position vectors of the points $P, Q, R,$ and $S$ with respect to $O$ are $\vec{p}, \vec{q}, \vec{r},$ and $\vec{s}$ respectively. The vector from point $M$ to point $N$ is given by $\overrightarrow{MN} = \vec{n} - \vec{m}$.

2. Use the given vector equation to find the position vector of S. The given relation is: $\overrightarrow{SP} + 5\; \overrightarrow{SQ} + 6\; \overrightarrow{SR} = \vec{0}$ We can express the vectors $\overrightarrow{SP}, \overrightarrow{SQ},$ and $\overrightarrow{SR}$ in terms of position vectors: $\overrightarrow{SP} = \vec{p} - \vec{s}$ $\overrightarrow{SQ} = \vec{q} - \vec{s}$ $\overrightarrow{SR} = \vec{r} - \vec{s}$

   Substituting these into the given equation: $(\vec{p} - \vec{s}) + 5(\vec{q} - \vec{s}) + 6(\vec{r} - \vec{s}) = \vec{0}$ $\vec{p} - \vec{s} + 5\vec{q} - 5\vec{s} + 6\vec{r} - 6\vec{s} = \vec{0}$ $\vec{p} + 5\vec{q} + 6\vec{r} - 12\vec{s} = \vec{0}$ Solving for $\vec{s}$: $12\vec{s} = \vec{p} + 5\vec{q} + 6\vec{r}$ $\vec{s} = \frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12}$

3. Find the position vectors of the mid-points E and F. $E$ is the mid-point of the side $PR$. Using the mid-point formula, the position vector of $E$, denoted by $\vec{e}$, is: $\vec{e} = \frac{\vec{p} + \vec{r}}{2}$ $F$ is the mid-point of the side $QR$. The position vector of $F$, denoted by $\vec{f}$, is: $\vec{f} = \frac{\vec{q} + \vec{r}}{2}$

4. Determine the vectors for the line segments EF and ES. The vector for the line segment $EF$ is $\overrightarrow{EF}$: $\overrightarrow{EF} = \vec{f} - \vec{e} = \left(\frac{\vec{q} + \vec{r}}{2}\right) - \left(\frac{\vec{p} + \vec{r}}{2}\right) = \frac{\vec{q} + \vec{r} - \vec{p} - \vec{r}}{2} = \frac{\vec{q} - \vec{p}}{2}$ The vector for the line segment $ES$ is $\overrightarrow{ES}$: $\overrightarrow{ES} = \vec{s} - \vec{e} = \left(\frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12}\right) - \left(\frac{\vec{p} + \vec{r}}{2}\right)$ To subtract, we find a common denominator: $\overrightarrow{ES} = \frac{\vec{p} + 5\vec{q} + 6\vec{r}}{12} - \frac{6(\vec{p} + \vec{r})}{12} = \frac{\vec{p} + 5\vec{q} + 6\vec{r} - 6\vec{p} - 6\vec{r}}{12}$ $\overrightarrow{ES} = \frac{-5\vec{p} + 5\vec{q}}{12} = \frac{5(\vec{q} - \vec{p})}{12}$

5. Calculate the lengths of the line segments EF and ES. The length of a line segment is the magnitude of its corresponding vector. Length of $EF = |\overrightarrow{EF}| = \left|\frac{\vec{q} - \vec{p}}{2}\right| = \frac{1}{2}|\vec{q} - \vec{p}|$ Length of $ES = |\overrightarrow{ES}| = \left|\frac{5(\vec{q} - \vec{p})}{12}\right| = \frac{5}{12}|\vec{q} - \vec{p}|$

6. Compute the required ratio. The ratio is $\frac{\text { length of the line segment } EF}{\text { length of the line segment } ES}$. $\frac{\text{length}(EF)}{\text{length}(ES)} = \frac{|\overrightarrow{EF}|}{|\overrightarrow{ES}|} = \frac{\frac{1}{2}|\vec{q} - \vec{p}|}{\frac{5}{12}|\vec{q} - \vec{p}|}$ Since $P$ and $Q$ are distinct points, $|\vec{q} - \vec{p}| \neq 0$, so we can cancel this term. $\frac{\text{length}(EF)}{\text{length}(ES)} = \frac{1/2}{5/12} = \frac{1}{2} \times \frac{12}{5} = \frac{12}{10} = \frac{6}{5} = 1.2$

Thus, the value of the ratio is 1.2.