(c)
Let $\frac{5z_{2}}{11z_{1}} = iy\ \ \ \Rightarrow \ \ \frac{z_{2}}{z_{1}} = \frac{11}{5}\ iy$
Now, $\left| \frac{2z_{1} + 3z_{2}}{2z_{1} - 3z_{2}} \right| = \left| \frac{2 + 3\frac{z_{2}}{z_{1}}}{2 - 3\frac{z_{2}}{z_{1}}} \right| = \left| \frac{2 + \frac{33}{5}iy}{2 - \frac{33}{5}iy} \right| = 1$