Solution:

To find the resulting temperature when two liquids A and B are mixed, we can use the principle of conservation of heat energy: the heat lost by the hotter liquid (A) equals the heat gained by the cooler liquid (B).

Given:
- Temperature of liquid A = 75°C
- Temperature of liquid B = 15°C
- Masses of A and B are in the ratio 2:3, so let the masses of A and B be \( 2m \) and \( 3m \), respectively.
- Specific heats of A and B are in the ratio 3:4, so let the specific heats of A and B be \( 3c \) and \( 4c \), respectively.

Let the final temperature of the mixture be \( T \).

Step 1: Heat lost by liquid A (hotter liquid):
\[
Q_A = \text{mass of A} \times \text{specific heat of A} \times (\text{initial temp of A} - T)
\]
\[
Q_A = 2m \times 3c \times (75 - T) = 6mc(75 - T)
\]

Step 2: Heat gained by liquid B (cooler liquid):
\[
Q_B = \text{mass of B} \times \text{specific heat of B} \times (T - \text{initial temp of B})
\]
\[
Q_B = 3m \times 4c \times (T - 15) = 12mc(T - 15)
\]

Step 3: Apply the conservation of heat:
Heat lost by A = Heat gained by B
\[
6mc(75 - T) = 12mc(T - 15)
\]

 Step 4: Simplify and solve for \( T \):
Cancel out \( mc \) from both sides:
\[
6(75 - T) = 12(T - 15)
\]
Expand both sides:
\[
450 - 6T = 12T - 180
\]
Combine like terms:
\[
450 + 180 = 12T + 6T
\]
\[
630 = 18T
\]
Solve for \( T \):
\[
T = \frac{630}{18} = 35 \, ^\circ\text{C}
\]

Final Answer:
The resulting temperature is 35°C.