A triangle with side lengths $16$, $18$, and $21$ has a circle with radius $6$ centered at each vertex. Find $n$ so that the total area inside the three circles but outside of the triangle is $n\pi$.
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1. **Calculate the area of the three circles:**
   Each circle has a radius of \(6\). The area of one circle is given by:
   \[
   \text{Area of one circle} = \pi r^2 = \pi \times 6^2 = 36\pi
   \]
   Since there are three circles, the total area of the three circles is:
   \[
   \text{Total area of three circles} = 3 \times 36\pi = 108\pi
   \]

2. **Calculate the area of the triangle:**
   The side lengths of the triangle are \(16\), \(18\), and \(21\). We can use Heron's formula to find the area of the triangle. First, we calculate the semi-perimeter \(s\):
   \[
   s = \frac{16 + 18 + 21}{2} = \frac{55}{2} = 27.5
   \]
   Using Heron's formula:
   \[
   \text{Area of the triangle} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{27.5(27.5-16)(27.5-18)(27.5-21)}
   \]
   \[
   = \sqrt{27.5 \times 11.5 \times 9.5 \times 6.5}
   \]
   \[
   = \sqrt{27.5 \times 11.5 \times 9.5 \times 6.5} \approx 126
   \]

3. **Calculate the area inside the three circles but outside the triangle:**
   The total area inside the three circles is \(108\pi\). We need to subtract the area of the triangle from this total. However, we must account for the fact that the triangle's vertices are inside the circles, and the angles at the vertices contribute to the area of the circles.

4. **Calculate the area of the sectors of the circles inside the triangle:**
   The sum of the angles in a triangle is \(180^\circ\). Each vertex of the triangle contributes a sector of the circle with an angle equal to the angle at that vertex. The total angle for the three sectors is \(180^\circ\), which is half of a full circle (\(360^\circ\)). Therefore, the area of these sectors is:
   \[
   \text{Area of the sectors} = \frac{1}{2} \times \text{Area of one circle} = \frac{1}{2} \times 36\pi = 18\pi
   \]
   Since there are three such sectors, the total area of the sectors is:
   \[
   \text{Total area of the sectors} = 3 \times 18\pi = 54\pi
   \]

5. **Calculate the area inside the three circles but outside the triangle:**
   The area inside the three circles but outside the triangle is the total area of the circles minus the area of the sectors:
   \[
   \text{Area inside the circles but outside the triangle} = 108\pi - 54\pi = 54\pi
   \]

The final answer is \(\boxed{54\pi}\)