Let the eccentricity $$e$$ of a hyperbola satisfy the equation $$6e^2 - 11e + 3 = 0$$. Its foci of the hyperbola are $$(3, 5)$$ and $$(3, -4)$$.then  the length of its latus rectum is :

The two foci are given as $$(3,5)$$ and $$(3,-4)$$. Since the $$x$$-coordinates are equal, the transverse axis of the hyperbola is vertical and the centre is the midpoint of the foci.

Centre $$\,(h,k)=\left(3,\dfrac{5+(-4)}{2}\right)=(3,0.5)$$.

The distance between the foci equals $$2c$$. Hence
$$2c=\sqrt{(3-3)^2+(5-(-4))^2}=|5-(-4)|=9 \;\Longrightarrow\; c=\dfrac{9}{2}=4.5$$.

The eccentricity $$e$$ satisfies the quadratic equation $$6e^2-11e+3=0$$.
Solving, $$e=\dfrac{11\pm\sqrt{11^2-4\cdot6\cdot3}}{12}= \dfrac{11\pm7}{12}$$, giving $$e_1=\dfrac{18}{12}=\dfrac32$$ and $$e_2=\dfrac{4}{12}=\dfrac13$$.

For a hyperbola $$e\gt1$$, so we take $$e=\dfrac32$$.

For a hyperbola $$e=\dfrac{c}{a}\; \Longrightarrow\; a=\dfrac{c}{e}= \dfrac{4.5}{1.5}=3 \quad\Rightarrow\quad a^2=9$$.

The relation between the semi-axes is $$c^2=a^2+b^2$$. Hence
$$b^2=c^2-a^2=(4.5)^2-9=20.25-9=11.25=\dfrac{45}{4}$$.

The length of the latus rectum of a hyperbola is $$L=\dfrac{2b^2}{a}$$.
$$L=\dfrac{2\left(\dfrac{45}{4}\right)}{3}= \dfrac{90}{4}\times\dfrac{1}{3}= \dfrac{90}{12}=7.5=\dfrac{15}{2}$$.

Therefore, the required length of the latus rectum is $$\dfrac{15}{2}$$.

Option C which is: $$\frac{15}{2}$$