Given that $a,b,c$ are integers with $abc = 60$, and that complex number $\omega \neq 1$ satisfies $\omega^3=1$, find the minimum possible value of $|a + b\omega + c\omega^2|$.


1. Given that \(a, b, c\) are integers with \(abc = 60\), and that the complex number \(\omega \neq 1\) satisfies \(\omega^3 = 1\), we need to find the minimum possible value of \(|a + b\omega + c\omega^2|\).

2. Since \(\omega\) is a cube root of unity, we have \(\omega^3 = 1\) and \(\omega \neq 1\). The cube roots of unity are \(1, \omega, \omega^2\), where \(\omega = e^{2\pi i / 3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\) and \(\omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\).

3. We can represent the points \(a\), \(b\omega\), and \(c\omega^2\) on the complex plane. The coordinates of these points are:
   - \(a\) at \((a, 0)\)
   - \(b\omega\) at \(\left(-\frac{b}{2}, \frac{b\sqrt{3}}{2}\right)\)
   - \(c\omega^2\) at \(\left(-\frac{c}{2}, -\frac{c\sqrt{3}}{2}\right)\)

4. We need to find the magnitude of the sum of these vectors:
   \[
   |a + b\omega + c\omega^2|
   \]

5. The sum of these vectors in the complex plane is:
   \[
   a + b\omega + c\omega^2 = a + b\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) + c\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)
   \]

6. Simplifying the expression:
   \[
   a + b\omega + c\omega^2 = a - \frac{b}{2} + i\frac{b\sqrt{3}}{2} - \frac{c}{2} - i\frac{c\sqrt{3}}{2}
   \]
   \[
   = a - \frac{b}{2} - \frac{c}{2} + i\left(\frac{b\sqrt{3}}{2} - \frac{c\sqrt{3}}{2}\right)
   \]
   \[
   = a - \frac{b+c}{2} + i\frac{\sqrt{3}}{2}(b - c)
   \]

7. The magnitude of this complex number is:
   \[
   \left|a - \frac{b+c}{2} + i\frac{\sqrt{3}}{2}(b - c)\right|
   \]
   \[
   = \sqrt{\left(a - \frac{b+c}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}(b - c)\right)^2}
   \]

8. Simplifying further:
   \[
   = \sqrt{\left(a - \frac{b+c}{2}\right)^2 + \frac{3}{4}(b - c)^2}
   \]

9. To minimize this expression, we need to find values of \(a, b, c\) such that \(abc = 60\) and the expression is minimized. We can try different combinations of \(a, b, c\) that satisfy \(abc = 60\).

10. Let's try \(a = 3\), \(b = 4\), \(c = 5\):
    \[
    abc = 3 \cdot 4 \cdot 5 = 60
    \]
    \[
    \left(a - \frac{b+c}{2}\right)^2 + \frac{3}{4}(b - c)^2 = \left(3 - \frac{4+5}{2}\right)^2 + \frac{3}{4}(4 - 5)^2
    \]
    \[
    = \left(3 - \frac{9}{2}\right)^2 + \frac{3}{4}(-1)^2
    \]
    \[
    = \left(3 - 4.5\right)^2 + \frac{3}{4}
    \]
    \[
    = (-1.5)^2 + \frac{3}{4}
    \]
    \[
    = 2.25 + 0.75
    \]
    \[
    = 3
    \]

11. Therefore, the minimum possible value of \(|a + b\omega + c\omega^2|\) is:
    \[
    \sqrt{3}
    \]

The final answer is \(\boxed{\sqrt{3}}\).