Consider the stress tensor
\begin{bmatrix} 0 & \tau & \tau \\ \tau & 0 & \tau\\ \tau & \tau & 0 \end{bmatrix}
To find the principal stresses with the characteristic equation we get
$ \sigma^3 - 3 \tau^2 \sigma - 2\tau^3 = 0$
and the Mohr's circle reduces to a point.
How to find the principal stresses and directions.
Hint:
You have an obvious direction $(1,1,1)$ associated with the principal stress $2\tau$. This tells you that the characteristic polynomial has $(\sigma-2\tau)$ as a factor. Can you then find the other factor? $$ \sigma^3 - 3 \tau^2 \sigma - 2\tau^3 = (\sigma-2\tau)(a\sigma^2+b\sigma+c)\Longrightarrow a=1, c=\tau^2,\ldots$$
It will be a quadratic expression, so finding the other two roots should not be a problem.
Then to find the other directions for the other two roots, this is two linear systems to solve: $$\begin{bmatrix} 0 & \tau & \tau \\ \tau & 0 & \tau\\ \tau & \tau & 0 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2\\ v_3 \end{bmatrix}=\sigma\begin{bmatrix} v_1 \\ v_2\\ v_3 \end{bmatrix}$$ replacing $\sigma$ by the roots that you have found.