The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$.  What is the least possible area of $\triangle ABC?$

$\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1$
1. The given equation is \((z+6)^8 = 81\). To solve for \(z\), we first take the eighth root of both sides:
   \[
   z + 6 = \sqrt[8]{81} e^{\frac{2\pi i k}{8}}, \quad k \in \{0, 1, 2, \ldots, 7\}
   \]
   Since \(81 = 3^4\), we have:
   \[
   \sqrt[8]{81} = \sqrt[8]{3^4} = 3^{\frac{4}{8}} = 3^{\frac{1}{2}} = \sqrt{3}
   \]
   Therefore, the solutions are:
   \[
   z + 6 = \sqrt{3} e^{\frac{2\pi i k}{8}}, \quad k \in \{0, 1, 2, \ldots, 7\}
   \]
   Solving for \(z\), we get:
   \[
   z = \sqrt{3} e^{\frac{2\pi i k}{8}} - 6
   \]

2. These solutions form the vertices of a regular octagon centered at \(-6\) with radius \(\sqrt{3}\) in the complex plane.

3. To find the least possible area of \(\triangle ABC\), we consider three consecutive vertices of the octagon. The vertices of the octagon are given by:
   \[
   z_k = \sqrt{3} e^{\frac{2\pi i k}{8}} - 6, \quad k \in \{0, 1, 2, \ldots, 7\}
   \]

4. The area of \(\triangle ABC\) is minimized when \(A\), \(B\), and \(C\) are consecutive vertices. Let's denote these vertices as \(z_0\), \(z_1\), and \(z_2\).

5. The angle between two consecutive vertices of the octagon is:
   \[
   \theta = \frac{2\pi}{8} = \frac{\pi}{4}
   \]

6. The area of \(\triangle ABC\) can be calculated using the formula for the area of a triangle formed by three points in the complex plane:
   \[
   \text{Area} = \frac{1}{2} \left| \text{Im} \left( z_1 \overline{z_2} + z_2 \overline{z_0} + z_0 \overline{z_1} \right) \right|
   \]

7. Since the vertices are consecutive, we can use the fact that the area of a triangle formed by three points on a circle is given by:
   \[
   \text{Area} = \frac{1}{2} R^2 \sin \theta
   \]
   where \(R\) is the radius of the circle and \(\theta\) is the angle subtended by the arc between two points.

8. For our octagon, \(R = \sqrt{3}\) and \(\theta = \frac{\pi}{4}\):
   \[
   \text{Area} = \frac{1}{2} (\sqrt{3})^2 \sin \left( \frac{\pi}{4} \right) = \frac{1}{2} \cdot 3 \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{4}
   \]

9. However, we need to consider the area of the triangle formed by three consecutive vertices, which involves subtracting the area of the overlapping triangles. The correct calculation involves:
   \[
   [ABC] = [OAB] + [OBC] - [OAC]
   \]
   where each of these areas is:
   \[
   [OAB] = \frac{\sqrt{3} \cdot \sqrt{3} \cdot \sin \left( \frac{\pi}{4} \right)}{2} = \frac{3 \cdot \frac{\sqrt{2}}{2}}{2} = \frac{3\sqrt{2}}{4}
   \]
   \[
   [OBC] = \frac{\sqrt{3} \cdot \sqrt{3} \cdot \sin \left( \frac{\pi}{4} \right)}{2} = \frac{3\sqrt{2}}{4}
   \]
   \[
   [OAC] = \frac{\sqrt{3} \cdot \sqrt{3} \cdot \sin \left( \frac{\pi}{2} \right)}{2} = \frac{3 \cdot 1}{2} = \frac{3}{2}
   \]

10. Therefore, the area of \(\triangle ABC\) is:
    \[
    [ABC] = \frac{3\sqrt{2}}{4} + \frac{3\sqrt{2}}{4} - \frac{3}{2} = \frac{3\sqrt{2}}{2} - \frac{3}{2}
    \]

The final answer is \(\boxed{\frac{3}{2}\sqrt{2} - \frac{3}{2}}\)