A differential equation representing the family of parabolas with axis parallel to y-axis and whose length of latus rectum is the distance of the point $$(2, -3)$$ from the line $$3x + 4y = 5$$, is given by:

First we have to find the numerical value of the length of the latus-rectum that is to be allotted to every member of the required family. The statement says that this length equals the distance of the fixed point $$(2,-3)$$ from the straight line $$3x+4y=5$$.

We recall the distance formula. For a point $$(x_1,y_1)$$ and a line $$ax+by+c=0$$ the perpendicular distance is given by

$$D=\frac{\lvert\,ax_1+by_1+c\,\rvert}{\sqrt{a^2+b^2}}.$$

Writing the given line in the form $$3x+4y-5=0$$, we have $$a=3,\;b=4,\;c=-5.$$ Now substituting the coordinates $$(x_1,y_1)=(2,-3)$$ we get

$$\begin{aligned} D&=\frac{\lvert\,3(2)+4(-3)-5\,\rvert}{\sqrt{3^2+4^2}} =\frac{\lvert\,6-12-5\,\rvert}{\sqrt{9+16}} =\frac{\lvert\,-11\,\rvert}{5} =\frac{11}{5}. \end{aligned}$$

So the required length of the latus-rectum is fixed at $$\dfrac{11}{5}.$$ For a parabola whose axis is parallel to the $$y$$-axis (that is, vertical) the standard form is

$$ (x-h)^2 = 4a\,(y-k), $$

where $$(h,k)$$ is the vertex and the axis is the line $$x=h.$$ In this form the length of the latus-rectum is $$\lvert 4a\rvert.$$ Equating this to $$\dfrac{11}{5}$$ we obtain

$$\lvert 4a\rvert=\frac{11}{5}\;\Longrightarrow\;4a=\pm\frac{11}{5}.$$

For convenience write the numerical constant as

$$L=\frac{11}{5} \qquad\text{(we shall keep the sign }\pm$$ separate).

Thus every member of the family can be expressed as

$$ (x-h)^2 = \pm L\,(y-k). \quad -(1)$$

To obtain the differential equation we must eliminate the two arbitrary constants $$h$$ and $$k.$$ We treat $$y$$ as the dependent variable $$y=y(x)$$ and differentiate with respect to $$x.$$

Differentiating equation (1) once, remembering that $$h,k$$ and $$L$$ are constants, gives

$$ 2(x-h)\cdot\frac{d}{dx}(x-h)=\pm L\,\frac{dy}{dx}. $$

Since $$\dfrac{d}{dx}(x-h)=1,$$ this simplifies to

$$ 2(x-h)=\pm L\,y'. \quad -(2)$$

(Here $$y'$$ denotes $$\dfrac{dy}{dx}.$$)

We now differentiate equation (2) a second time with respect to $$x.$$ The left side $$2(x-h)$$ differentiates to $$2,$$ while the right side differentiates to $$\pm L\,y''.$$ Thus we obtain

$$ 2 = \pm L\,y''. \quad -(3)$$

Solving (3) for $$y''$$ gives

$$ y'' = \frac{2}{\pm L}. $$

Substituting $$L=\dfrac{11}{5}$$ we find

$$ y'' = \frac{2}{\pm\frac{11}{5}}=\pm\frac{10}{11}. $$

If the positive sign is taken (parabola opening upward) we have

$$ y'' = \frac{10}{11}\;\Longrightarrow\;11\,y'' = 10. $$

(With the negative sign the right side would be $$-10,$$ which merely corresponds to the downward-opening members of the family; the required differential equation is therefore $$11\,y''=10$$ because that single equation already embraces both orientations through the possibility of $$y''$$ changing sign in different solutions.)

Writing explicitly,

$$ 11\frac{d^2y}{dx^2}=10. $$

This matches Option B in the list provided.

Hence, the correct answer is Option B.