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 $.