1. **State the problem:** Given that $\csc A = \frac{13}{12}$, find the value of $$\frac{2 \sin A - 3 \cos A}{4 \sin A - 9 \cos A}.$$\n\n2. **Recall the definitions and formulas:** \n- $\csc A = \frac{1}{\sin A}$, so $\sin A = \frac{1}{\csc A}$.\n- Use the Pythagorean identity: $\sin^2 A + \cos^2 A = 1$ to find $\cos A$.\n\n3. **Find $\sin A$:**\n$$\sin A = \frac{1}{\csc A} = \frac{1}{\frac{13}{12}} = \frac{12}{13}.$$\n\n4. **Find $\cos A$ using the Pythagorean identity:**\n$$\cos A = \pm \sqrt{1 - \sin^2 A} = \pm \sqrt{1 - \left(\frac{12}{13}\right)^2} = \pm \sqrt{1 - \frac{144}{169}} = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13}.$$\n\n5. **Determine the sign of $\cos A$:**\nSince $\csc A = \frac{13}{12} > 0$, $\sin A > 0$, so angle $A$ is in either the first or second quadrant.\n- In the first quadrant, $\cos A > 0$.\n- In the second quadrant, $\cos A < 0$.\nWithout additional information, we consider both cases.\n\n6. **Calculate the expression for both cases:**\n- For $\cos A = \frac{5}{13}$:\n$$\frac{2 \sin A - 3 \cos A}{4 \sin A - 9 \cos A} = \frac{2 \times \frac{12}{13} - 3 \times \frac{5}{13}}{4 \times \frac{12}{13} - 9 \times \frac{5}{13}} = \frac{\frac{24}{13} - \frac{15}{13}}{\frac{48}{13} - \frac{45}{13}} = \frac{\frac{9}{13}}{\frac{3}{13}} = 3.$$\n- For $\cos A = -\frac{5}{13}$:\n$$\frac{2 \sin A - 3 \cos A}{4 \sin A - 9 \cos A} = \frac{2 \times \frac{12}{13} - 3 \times \left(-\frac{5}{13}\right)}{4 \times \frac{12}{13} - 9 \times \left(-\frac{5}{13}\right)} = \frac{\frac{24}{13} + \frac{15}{13}}{\frac{48}{13} + \frac{45}{13}} = \frac{\frac{39}{13}}{\frac{93}{13}} = \frac{39}{93} = \frac{13}{31}.$$\n\n7. **Final answer:**\nThe value of the expression is either $3$ or $\frac{13}{31}$ depending on the quadrant of angle $A$.