Three distinct integers $a, b$ and $c$ satisfy the following three conditions:

- $a b c=17955$,
- $a, b$ and $c$ form an arithmetic sequence in that order, and
- $(3 a+b),(3 b+c)$, and $(3 c+a)$ form a geometric sequence in that order.

What is the value of $a+b+c$ ?

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, $3,5,7$ is an arithmetic sequence with three terms.

A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant. For example, $3,6,12$ is a geometric sequence with three terms.)
(A) -63
(B) -42
(C) -68229
(D) -48
(E) 81
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 $a b c=17955$ 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

$3 a+b=3(b-d)+b=4 b-3 d \quad 3 b+c=3 b+(b+d)=4 b+d \quad 3 c+a=3(b+d)+(b-d)=4 b+2 d$

Since $3 a+b, 3 b+c$ and $3 c+a$ form a geometric sequence in this order, then

$$
\begin{aligned}
\frac{3 b+c}{3 a+b} & =\frac{3 c+a}{3 b+c} \\
(3 b+c)^{2} & =(3 a+b)(3 c+a) \\
(4 b+d)^{2} & =(4 b-3 d)(4 b+2 d) \\
16 b^{2}+8 b d+d^{2} & =16 b^{2}-4 b d-6 d^{2} \\
12 b d & =-7 d^{2} \\
12 b & =-7 d \quad(\text { since } d \neq 0) \\
d & =-\frac{12}{7} b
\end{aligned}
$$

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 $a b c=17955$, 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 $3 a+b=-192$, $3 b+c=-48$, and $3 c+a=-12$ form a geometric sequence (with common ratio $\frac{1}{4}$ ). Therefore, $a+b+c=(-57)+(-21)+15=-63$.