jmc

In the arithmetic sequence $a,$ $b,$ $c,$ let $d$ be the common difference, so $a=b-d$ and $c=b+d.$ Then $$\begin{array}{rl}3a+b& =3(b-d)+b=4b-3d,\\ 3b+c& =3b+b+d=4b+d,\\ 3c+a& =3(b+d)+(b-d)=4b+2d,\end{array}$$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 $a+b+c=\boxed{-63}.$