Define $F:\mathbb{C}^{3}\rightarrow\mathbb{R}\times\mathbb{C}$ by $F(z_{1},z_{2},z_{3})=(a,c)$ with $2a=|z_{1}|^{2}-|z_{2}|^{2}$ and

$$c=\begin{cases}z_{3}&a=z_{1}=z_{2}=0\\ z_{3}-\bar{z}_{1}\bar{z}_{2}/|z_{1}|&a\geq 0,z_{1}\not=0\\ z_{3}-\bar{z}_{1}\bar{z}_{2}/|z_{2}|&a<0.\end{cases}$$

It is easy to see that if $a\not=0$, then $F^{-1}(a,c)$ is homeomorphic to $\mathbb{R}^{2}\times S^{1}$, while if $a=0$, then $F^{-1}(a,c)$ is a cone over $T^{2}$: essentially, one copy of $S^{1}$ in $\mathbb{R}^{2}\times S^{1}$ collapses to a point. In addition, all fibres of this map are special Lagrangian, and it is obviously only piecewise smooth. The discriminant locus is the entire plane given by $a=0$.