Let $ABCD$ be a square of side $a$. On side $AD$ consider points $E$ and $Z$ such that $DE=a/3$ and $AZ=a/4$. If the lines $BZ$ and $CE$ intersect at point $H$, calculate the area of the triangle $BCH$ in terms of $a$.
1. **Define the coordinates of the points:**
   - Let \( A = (0, 0) \), \( B = (a, 0) \), \( C = (a, a) \), and \( D = (0, a) \).
   - Point \( E \) on \( AD \) such that \( DE = \frac{a}{3} \) implies \( E = (0, \frac{2a}{3}) \).
   - Point \( Z \) on \( AD \) such that \( AZ = \frac{a}{4} \) implies \( Z = (0, \frac{a}{4}) \).

2. **Find the equations of lines \( BZ \) and \( CE \):**
   - The slope of \( BZ \) is \( \frac{\frac{a}{4} - 0}{0 - a} = -\frac{1}{4a} \), so the equation of \( BZ \) is:
     \[
     y = -\frac{1}{4}x + \frac{a}{4}
     \]
   - The slope of \( CE \) is \( \frac{\frac{2a}{3} - a}{0 - a} = \frac{1}{3} \), so the equation of \( CE \) is:
     \[
     y = \frac{1}{3}x + \frac{2a}{3}
     \]

3. **Find the intersection point \( H \) of \( BZ \) and \( CE \):**
   - Set the equations equal to each other:
     \[
     -\frac{1}{4}x + \frac{a}{4} = \frac{1}{3}x + \frac{2a}{3}
     \]
   - Solve for \( x \):
     \[
     -\frac{1}{4}x - \frac{1}{3}x = \frac{2a}{3} - \frac{a}{4}
     \]
     \[
     -\frac{3x}{12} - \frac{4x}{12} = \frac{8a - 3a}{12}
     \]
     \[
     -\frac{7x}{12} = \frac{5a}{12}
     \]
     \[
     x = -\frac{5a}{7}
     \]
   - Substitute \( x \) back into one of the line equations to find \( y \):
     \[
     y = \frac{1}{3} \left( -\frac{5a}{7} \right) + \frac{2a}{3}
     \]
     \[
     y = -\frac{5a}{21} + \frac{14a}{21}
     \]
     \[
     y = \frac{9a}{21} = \frac{3a}{7}
     \]
   - Thus, \( H = \left( -\frac{5a}{7}, \frac{3a}{7} \right) \).

4. **Calculate the area of triangle \( BCH \):**
   - Use the determinant formula for the area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
     \[
     \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
     \]
   - For triangle \( BCH \):
     \[
     \text{Area} = \frac{1}{2} \left| a \left( a - \frac{3a}{7} \right) + a \left( \frac{3a}{7} - 0 \right) + \left( -\frac{5a}{7} \right) (0 - a) \right|
     \]
     \[
     = \frac{1}{2} \left| a \left( \frac{4a}{7} \right) + a \left( \frac{3a}{7} \right) + \left( -\frac{5a}{7} \right) (-a) \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{4a^2}{7} + \frac{3a^2}{7} + \frac{5a^2}{7} \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{12a^2}{7} \right|
     \]
     \[
     = \frac{6a^2}{7}
     \]

The final answer is \(\boxed{\frac{6a^2}{7}}\)