Equilateral triangle $ABC$ has side length $6$. Circles with centers at $A$, $B$, and $C$ are drawn such that their respective radii $r_A$, $r_B$, and $r_C$ form an arithmetic sequence with $r_A<r_B<r_C$. If the shortest distance between circles $A$ and $B$ is $3.5$, and the shortest distance between circles $A$ and $C$ is $3$, then what is the area of the shaded region? Express your answer in terms of pi.

[asy]
size(8cm);
draw((0,0)--(6,0)--6*dir(60)--cycle);
draw(circle((0,0),1));
draw(circle(6*dir(60),1.5));
draw(circle((6,0),2));
filldraw((0,0)--arc((0,0),1,0,60)--cycle, grey);
filldraw(6*dir(60)--arc(6*dir(60),1.5,240,300)--cycle, grey);
filldraw((6,0)--arc((6,0),2,120,180)--cycle, grey);
label("$A$",(0,0),SW);
label("$B$",6*dir(60),N);
label("$C$",(6,0),SE);
[/asy]
1. Let the radii of the circles centered at \( A \), \( B \), and \( C \) be \( r_A = a \), \( r_B = b \), and \( r_C = c \) respectively. Given that \( r_A < r_B < r_C \) and they form an arithmetic sequence, we can write:
   \[
   b = a + d \quad \text{and} \quad c = a + 2d
   \]
   for some common difference \( d \).

2. The shortest distance between the circles centered at \( A \) and \( B \) is given as \( 3.5 \). Since the side length of the equilateral triangle is \( 6 \), we have:
   \[
   6 - (a + b) = 3.5
   \]
   Substituting \( b = a + d \) into the equation, we get:
   \[
   6 - (a + (a + d)) = 3.5 \implies 6 - (2a + d) = 3.5 \implies 2a + d = 2.5 \quad \text{(1)}
   \]

3. The shortest distance between the circles centered at \( A \) and \( C \) is given as \( 3 \). Similarly, we have:
   \[
   6 - (a + c) = 3
   \]
   Substituting \( c = a + 2d \) into the equation, we get:
   \[
   6 - (a + (a + 2d)) = 3 \implies 6 - (2a + 2d) = 3 \implies 2a + 2d = 3 \quad \text{(2)}
   \]

4. From equation (2), we can solve for \( d \):
   \[
   2a + 2d = 3 \implies a + d = 1.5 \quad \text{(3)}
   \]

5. Subtract equation (3) from equation (1):
   \[
   (2a + d) - (a + d) = 2.5 - 1.5 \implies a = 1
   \]

6. Substitute \( a = 1 \) back into equation (3):
   \[
   1 + d = 1.5 \implies d = 0.5
   \]

7. Now, we can find \( b \) and \( c \):
   \[
   b = a + d = 1 + 0.5 = 1.5
   \]
   \[
   c = a + 2d = 1 + 2(0.5) = 2
   \]

8. The area of the shaded region is \(\frac{1}{6}\) of the area of each circle. The areas of the circles are:
   \[
   \text{Area of circle at } A = \pi a^2 = \pi (1)^2 = \pi
   \]
   \[
   \text{Area of circle at } B = \pi b^2 = \pi (1.5)^2 = 2.25\pi
   \]
   \[
   \text{Area of circle at } C = \pi c^2 = \pi (2)^2 = 4\pi
   \]

9. The total area of the shaded regions is:
   \[
   \frac{1}{6} \left( \pi + 2.25\pi + 4\pi \right) = \frac{1}{6} \left( 7.25\pi \right) = \frac{7.25\pi}{6} = \frac{29\pi}{24}
   \]

The final answer is \(\boxed{\frac{29\pi}{24}}\).