Step 1: Rewrite the equation in geometric form.
The given equation is $(x-2)^2 + (y-3)^2 = \dfrac{1}{25}(3x-4y+7)^2$. Taking positive square roots: \[\sqrt{(x-2)^2+(y-3)^2} = \frac{|3x-4y+7|}{5}\]
Step 2: Recognise the parabola definition.
The left side is the distance from $(x,y)$ to the point $(2,3)$. The right side: since $\sqrt{3^2+(-4)^2} = 5$, we have $\dfrac{|3x-4y+7|}{5}$ = perpendicular distance from $(x,y)$ to the line $3x-4y+7=0$. This equation says: distance from point to $(2,3)$ = distance from point to the line $3x-4y+7=0$. By the definition of a parabola, the focus is $S(2,3)$ and the directrix is $3x-4y+7=0$.
Step 3: Compute the distance from the focus to the directrix.
\[d = \frac{|3(2) - 4(3) + 7|}{\sqrt{9+16}} = \frac{|6-12+7|}{5} = \frac{1}{5}\]
Step 4: Find the semi-latus-rectum parameter $a$.
For a parabola, the distance from the focus to the directrix equals $2a$: \[2a = \frac{1}{5} \implies a = \frac{1}{10}\]
Step 5: Apply the latus rectum formula.
The length of the latus rectum (the chord through the focus perpendicular to the axis) is $4a$: \[4a = 4 \times \frac{1}{10} = \frac{2}{5}\]
Step 6: Confirm by checking units and formula.
The latus rectum length $= 4a = \dfrac{2}{5}$. This is a positive number, as expected for a length.
Step 7: State the final answer.
\[ \boxed{\frac{2}{5}} \]