Let the eccentricity of a hyperbola satisfy the equation \( 6e^{2} - 11e + 3 = 0 \). If the foci of the hyperbola are \( (3, 5) \) and \( (3, -4) \), then the length of its latus rectum is:
\( \frac{11}{3} \)
\( \frac{17}{3} \)
\( \frac{15}{2} \)
\( \frac{17}{2} \)