In the arithmetic sequence $ a, $ $ b, $ $ c, $ let $ d $ be the common difference, so $ a = b - d $ and $ c = b + d $. Then \begin{align*} 3a + b &= 3(b - d) + b = 4b - 3d, \\ 3b + c &= 3b + b + d = 4b + d, \\ 3c + a &= 3(b + d) + (b - d) = 4b + 2d, \end{align*}so \[(4b + d)^2 = (4b - 3d)(4b + 2d).\]This simplifies to $ 12bd + 7d^2 = d(12b + 7d) = 0 $. If $ d = 0, $ then $ a = b = c, $ so $ a^3 = 17955 $. Since 17955 is not a perfect cube, $ 12b + 7d = 0, $ so $ d = -\frac{12}{7} b $. Then $ a = b - d = \frac{19}{7} b $ and $ c = b + d = -\frac{5}{7} b $. Substituting into $ abc = 17955, $ we get \[\frac{19}{7} b \cdot b \cdot \left( -\frac{5}{7} b \right) = 17955.\]Then $ b^3 = -9261, $ so $ b = -21 $. Hence, $ a = -57 $ and $ c = 15, $ so $ a + b + c = \boxed{-63} $.