Given is a acute angled triangle $ABC$. The lengths of the altitudes from $A, B$ and $C$ are successively $h_A, h_B$ and $h_C$. Inside the triangle is a point $P$. The distance from $P$ to $BC$ is $1/3 h_A$ and the distance from $P$ to $AC$ is $1/4 h_B$. Express the distance from $P$ to $AB$ in terms of $h_C$.
1. Let the area of triangle \( \triangle ABC \) be denoted as \( A_{\triangle ABC} \). The area can be expressed in terms of the altitudes and the corresponding bases. For example, using the altitude \( h_A \) from vertex \( A \) to side \( BC \), we have:
   \[
   A_{\triangle ABC} = \frac{1}{2} \cdot BC \cdot h_A
   \]

2. Similarly, the areas of the sub-triangles \( \triangle PBC \), \( \triangle PAC \), and \( \triangle PAB \) can be expressed using the distances from point \( P \) to the sides \( BC \), \( AC \), and \( AB \) respectively. Let these distances be \( h_1 \), \( h_2 \), and \( h_3 \).

3. Given that the distance from \( P \) to \( BC \) is \( \frac{1}{3} h_A \), we have:
   \[
   h_1 = \frac{1}{3} h_A
   \]
   The area of \( \triangle PBC \) is:
   \[
   A_{\triangle PBC} = \frac{1}{2} \cdot BC \cdot h_1 = \frac{1}{2} \cdot BC \cdot \frac{1}{3} h_A = \frac{1}{3} A_{\triangle ABC}
   \]
   Therefore:
   \[
   \frac{A_{\triangle PBC}}{A_{\triangle ABC}} = \frac{1}{3}
   \]

4. Given that the distance from \( P \) to \( AC \) is \( \frac{1}{4} h_B \), we have:
   \[
   h_2 = \frac{1}{4} h_B
   \]
   The area of \( \triangle PAC \) is:
   \[
   A_{\triangle PAC} = \frac{1}{2} \cdot AC \cdot h_2 = \frac{1}{2} \cdot AC \cdot \frac{1}{4} h_B = \frac{1}{4} A_{\triangle ABC}
   \]
   Therefore:
   \[
   \frac{A_{\triangle PAC}}{A_{\triangle ABC}} = \frac{1}{4}
   \]

5. The sum of the areas of the sub-triangles \( \triangle PBC \), \( \triangle PAC \), and \( \triangle PAB \) must equal the area of \( \triangle ABC \):
   \[
   \frac{A_{\triangle PBC}}{A_{\triangle ABC}} + \frac{A_{\triangle PAC}}{A_{\triangle ABC}} + \frac{A_{\triangle PAB}}{A_{\triangle ABC}} = 1
   \]
   Substituting the known ratios:
   \[
   \frac{1}{3} + \frac{1}{4} + \frac{A_{\triangle PAB}}{A_{\triangle ABC}} = 1
   \]

6. Solving for \( \frac{A_{\triangle PAB}}{A_{\triangle ABC}} \):
   \[
   \frac{A_{\triangle PAB}}{A_{\triangle ABC}} = 1 - \frac{1}{3} - \frac{1}{4} = 1 - \frac{4}{12} - \frac{3}{12} = 1 - \frac{7}{12} = \frac{5}{12}
   \]

7. Since \( \frac{A_{\triangle PAB}}{A_{\triangle ABC}} = \frac{h_3}{h_C} \), we have:
   \[
   \frac{h_3}{h_C} = \frac{5}{12}
   \]
   Therefore:
   \[
   h_3 = \frac{5}{12} h_C
   \]

The final answer is \( \boxed{ \frac{5}{12} h_C } \)