Ratio of Segment Lengths Formed by Vectors in Triangle

Let the position vectors of the points $P, Q, R,$ and $S$ with respect to the point $S$ as the origin be $\vec{p}, \vec{q}, \vec{r},$ and $\vec{0}$, respectively.

Given the relation:

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

In terms of position vectors relative to $S$:

$$\vec{p} + 5\vec{q} + 6\vec{r} = \vec{0} \implies \vec{p} = -5\vec{q} - 6\vec{r}$$

Since $E$ is the midpoint of $PR$, its position vector $\vec{e}$ is:

$$\vec{e} = \frac{\vec{p} + \vec{r}}{2} = \frac{(-5\vec{q} - 6\vec{r}) + \vec{r}}{2} = -\frac{5}{2}(\vec{q} + \vec{r})$$

Since $F$ is the midpoint of $QR$, its position vector $\vec{f}$ is:

$$\vec{f} = \frac{\vec{q} + \vec{r}}{2}$$

Now, the vector representing the line segment $EF$ is:

$$\vec{EF} = \vec{f} - \vec{e} = \frac{\vec{q} + \vec{r}}{2} - \left(-\frac{5}{2}(\vec{q} + \vec{r})\right) = 3(\vec{q} + \vec{r})$$

The vector representing the line segment $ES$ is:

$$\vec{ES} = \vec{0} - \vec{e} = \frac{5}{2}(\vec{q} + \vec{r})$$

Therefore, the ratio of their lengths is:

$$\frac{\text{length of the line segment } EF}{\text{length of the line segment } ES} = \frac{\|\vec{EF}\|}{\|\vec{ES}\|} = \frac{3\|\vec{q} + \vec{r}\|}{\frac{5}{2}\|\vec{q} + \vec{r}\|} = \frac{6}{5} = 1.2$$