An algebra problem by Aira Thalca

Let the integers be $a,b=a+d,c=a+2d$ (since they form an AP). Since $abc=17955\\ \implies a(a+d)(a+2d)=17955...(1)$

$(3a + b), (3b + c), (3c + a)$ form a geometric sequence $\implies (3b+c)^2=(3a+b)(3c+a)$ $\implies (4a+5d)^2=(4a+d)(4a+6d)$

$\implies 19d^2+12ad=0\implies a=\frac{-19d}{12}$ Substituting in $(1)$ we get:

$\frac{19d}{12}\cdot \frac{7d}{12}\cdot \frac{5d}{12}=17955=3^3\cdot 19\cdot 7\cdot 5$

$\implies d=36\implies a=\dfrac{-19d}{12}=-57$

The sum of these terms is: $a+b+c=3a+3d=3(36+(-57))=\boxed{\color{#69047E}{-63}}$