Since $a, b$ and $c$ form an arithmetic sequence in this order, then $a=b-d$ and $c=b+d$ for some real number $d$. We note that $d \neq 0$, since otherwise we would have $a=b=c$ and then $abc=17955$ would tell us that $b^{3}=17955$ or $b=\sqrt[3]{17955}$, which is not an integer. Writing the terms of the geometric sequence in terms of $b$ and $d$, we have $3a+b=3(b-d)+b=4b-3d \quad 3b+c=3b+(b+d)=4b+d \quad 3c+a=3(b+d)+(b-d)=4b+2d$. Since $3a+b, 3b+c$ and $3c+a$ form a geometric sequence in this order, then $\frac{3b+c}{3a+b} =\frac{3c+a}{3b+c}$. Simplifying, we get $12b=-7d$ and $d=-\frac{12}{7}b$. Therefore, $a=b-d=b-\left(-\frac{12}{7}b\right)=\frac{19}{7}b$ and $c=b+d=b+\left(-\frac{12}{7}b\right)=-\frac{5}{7}b$. Since $abc=17955$, then $\left(\frac{19}{7}b\right)(b)\left(-\frac{5}{7}b\right)=17955$ or $-\frac{95}{49}b^{3}=17955$ or $b^{3}=-9261$ and so $b=-21$. Thus, $a=\frac{19}{7}b=\frac{19}{7}(-21)=-57$ and $c=-\frac{5}{7}b=-\frac{5}{7}(-21)=15$. We can check that $a=-57, b=-21$ and $c=15$ have a product of 17955, that $-57,-21,15$ is indeed an arithmetic sequence (with common difference 36), and that $3a+b=-192$, $3b+c=-48$, and $3c+a=-12$ form a geometric sequence (with common ratio $\frac{1}{4}$). Therefore, $a+b+c=(-57)+(-21)+15=-63$.