Let $a,$ $b,$ $c$ be nonzero real numbers.  Find the maximum angle between the vectors $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$ and $\begin{pmatrix} b \\ c \\ a \end{pmatrix},$ in degrees.
Let $\theta$ be the angle between the two vectors.  Then
\[\cos \theta = \frac{\begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} b \\ c \\ a \end{pmatrix}}{\left\| \begin{pmatrix} a \\ b \\ c \end{pmatrix} \right\| \left\|\begin{pmatrix} b \\ c \\ a \end{pmatrix} \right\|} = \frac{ab + ac + bc}{a^2 + b^2 + c^2}.\]We can say that
\[(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \ge 0,\]so $2(ab + ac + bc) \ge -(a^2 + b^2 + c^2).$  Since $a,$ $b,$ and $c$ are nonzero, $a^2 + b^2 + c^2 > 0,$ so
\[\frac{2(ab + ac + bc)}{a^2 + b^2 + c^2} \ge -1.\]Hence,
\[\cos \theta = \frac{ab + ac + bc}{a^2 + b^2 + c^2} \ge -\frac{1}{2}.\]The largest angle $\theta$ that satisfies this is $\boxed{120^\circ}.$  Equality occurs for any nonzero real numbers $a,$ $b,$ $c$ that satisfy $a + b + c = 0.$