Determine the value of the expression $x^2 + y^2 + z^2$, 

if $x + y + z = 13$ , $xyz= 72$ and $\frac1x + \frac1y + \frac1z = \frac34$.
1. Given the equations:
   \[
   x + y + z = 13,
   \]
   \[
   xyz = 72,
   \]
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{3}{4},
   \]
   we need to determine the value of \( x^2 + y^2 + z^2 \).

2. First, we use the identity for the sum of the reciprocals:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{xy + yz + zx}{xyz}.
   \]
   Substituting the given values, we have:
   \[
   \frac{xy + yz + zx}{72} = \frac{3}{4}.
   \]

3. Solving for \( xy + yz + zx \):
   \[
   xy + yz + zx = 72 \cdot \frac{3}{4} = 54.
   \]

4. Next, we use the identity for the sum of squares:
   \[
   x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx).
   \]
   Substituting the known values:
   \[
   x^2 + y^2 + z^2 = 13^2 - 2 \cdot 54.
   \]

5. Calculating the right-hand side:
   \[
   13^2 = 169,
   \]
   \[
   2 \cdot 54 = 108,
   \]
   \[
   x^2 + y^2 + z^2 = 169 - 108 = 61.
   \]

6. Therefore, the value of the expression \( x^2 + y^2 + z^2 \) is:
   \[
   \boxed{61}.
   \]