A triangle has sides of lengths 5, 6, 7. What is 60 times the square of the radius of the inscribed circle?
1. **Calculate the semiperimeter \( s \) of the triangle:**
   The semiperimeter \( s \) is given by:
   \[
   s = \frac{a + b + c}{2}
   \]
   where \( a = 5 \), \( b = 6 \), and \( c = 7 \). Therefore,
   \[
   s = \frac{5 + 6 + 7}{2} = \frac{18}{2} = 9
   \]

2. **Calculate the area \( A \) of the triangle using Heron's formula:**
   Heron's formula states that the area \( A \) of a triangle with sides \( a \), \( b \), and \( c \) is:
   \[
   A = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   Substituting the values:
   \[
   A = \sqrt{9(9-5)(9-6)(9-7)} = \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = \sqrt{216} = 6\sqrt{6}
   \]

3. **Relate the area to the inradius \( r \):**
   The area \( A \) of a triangle can also be expressed in terms of the inradius \( r \) and the semiperimeter \( s \) as:
   \[
   A = r \cdot s
   \]
   Given \( A = 6\sqrt{6} \) and \( s = 9 \), we can solve for \( r \):
   \[
   6\sqrt{6} = r \cdot 9 \Rightarrow r = \frac{6\sqrt{6}}{9} = \frac{2\sqrt{6}}{3}
   \]

4. **Calculate \( 60r^2 \):**
   First, find \( r^2 \):
   \[
   r^2 = \left( \frac{2\sqrt{6}}{3} \right)^2 = \frac{4 \cdot 6}{9} = \frac{24}{9} = \frac{8}{3}
   \]
   Then, multiply by 60:
   \[
   60r^2 = 60 \cdot \frac{8}{3} = 20 \cdot 8 = 160
   \]

The final answer is \( \boxed{160} \).