Three planets orbit a star circularly in the same plane.  Each moves in the same direction and moves at [constant](https://artofproblemsolving.com/wiki/index.php/Constant) speed.  Their periods are 60, 84, and 140 years.  The three planets and the star are currently [collinear](https://artofproblemsolving.com/wiki/index.php/Collinear).  What is the fewest number of years from now that they will all be collinear again?
Denote the planets $A, B, C$ respectively.  Let $a(t), b(t), c(t)$ denote the angle which each of the respective planets makes with its initial position after $t$ years.  These are given by $a(t) = \frac{t \pi}{30}$, $b(t) = \frac{t \pi}{42}$, $c(t) = \frac{t \pi}{70}$.
In order for the planets and the central star to be collinear, $a(t)$, $b(t)$, and $c(t)$ must differ by a multiple of $\pi$.  Note that $a(t) - b(t) = \frac{t \pi}{105}$ and $b(t) - c(t) = \frac{t \pi}{105}$, so $a(t) - c(t) = \frac{ 2 t \pi}{105}$.  These are simultaneously multiples of $\pi$ exactly when $t$ is a multiple of $105$, so the planets and the star will next be collinear in $\boxed{105}$ years.