I love it when the planets come together

The Xiddlerian astronomers have identified three non-Keplerian planets that circularly orbit a neighboring star. Planet A is three astronomical units away from its star and completes its orbit in three years. Planet B is four astronomical units away from the star and completes its orbit in four years. Finally, Planet C is five astronomical units away from the star and completes its orbit in five years.

They report their findings to Xiddler’s Grand Minister, along with the auspicious news that all three planets are currently lined up with their star. However, the Grand Minister is far more interested in the three planets than the star and wants to know how long it will be until the planets are next aligned. How many years will it be until the three planets are again collinear (not necessarily including the star)?

Let's denote the positions of planets $A,$ $B$ and $C$ at time $t$ (in years) as \begin{align*}A(t) &= (3\cos \frac{2\pi t}{3}, 3\sin \frac{2\pi t}{3}),\\ B(t) &= (4\cos \frac{\pi t}{2}, 4\sin \frac{\pi t}{2}), \\ C(t) &= (5\cos \frac{2\pi t}{5}, 5 \sin \frac{2\pi t}{5}),\end{align*} respectively. In order to test for collinearity of $A(t)$, $B(t)$ and $C(t)$ we will want to test whether the area of the triangle with ABC has zero area. Given two vectors $\vec{u}, \vec{v} \in \mathbb{R}^2$ the area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$ is given by the norm of their cross product $\| \vec{u} \times \vec{v} \|$ therefore we can calculate the area of the triangle spanned by $A(t),$ $B(t)$ and $C(t)$ is given by \begin{align*}S(t) &= \frac{1}{2} \| (B(t) - A(t)) \times (C(t) - A(t)) \|\\ &= \frac{1}{2} \text{abs} \left( \begin{vmatrix} 4\cos \frac{\pi t}{2} - 3\cos \frac{2\pi t}{3} & 4\sin \frac{\pi t}{2} - 3 \sin \frac{2\pi t}{3} \\ 5\cos \frac{2\pi t}{5} - 3 \cos \frac{2\pi t}{3} & 5 \sin \frac{2\pi t}{5} - 3 \sin \frac{2\pi t}{3} \end{vmatrix} \right)\\ &= \frac{1}{2} \left|(4 \cos \frac{\pi t}{2} - 3 \cos \frac{2\pi t}{3}) (5 \sin \frac{2\pi t}{5} - 3 \sin \frac{2\pi t}{3} ) \right.\\ & \quad\quad\quad\quad \left. - (5 \cos \frac{2\pi t}{5} - 3 \cos \frac{2\pi t}{3}) (4 \sin \frac{\pi t}{2} - 3 \sin \frac{2\pi t}{3}) \right|\\ &= \frac{1}{2} \left|\,\, 20 \cos \frac{\pi t}{2} \sin \frac{2\pi t}{5} - 12 \cos \frac{\pi t}{2} \sin \frac{2\pi t}{3} - 15 \cos \frac{2\pi t}{3} \sin \frac{2\pi t}{5} + 9 \cos \frac{2\pi t}{3} \sin \frac{2\pi t}{3} \right. \\ & \quad\quad \left. - 20 \cos \frac{2\pi t}{5} \sin \frac{\pi t}{2} + 12 \cos \frac{2\pi t}{3} \sin \frac{\pi t}{2} + 15 \cos \frac{2\pi t}{5} \sin \frac{2\pi t}{3} - 9 \cos \frac{2\pi t}{3} \sin \frac{2\pi t}{3} \right| \\ &= \frac{1}{2} \left| -20 \sin \frac{\pi t}{10} - 12 \sin \frac{\pi t}{6} + 15 \sin \frac{4\pi t}{15} \right| \end{align*}

Obviously in the context of area, we would want to keep the absolute value, though since we are trying to find zeros of this trigonometric function, it might be easier to just deal with the signed version $$\tilde{S}(t) = \frac{1}{2} \left( 15 \sin \frac{4\pi t}{15} - 20 \sin \frac{\pi t}{10} - 12 \sin \frac{\pi t}{6} \right).$$ The signed area is shown in the figure below for the first full period (until $t = 60$ years).

We see from eyeballing it, as all good Xiddlerian astronomers are wont to do, that the first zero of $\tilde{S}$ is between $t = 5$ and $t = 10$ years, with $\tilde{S}(5) = -19.4952$ and $\tilde{S}(10) = 11.6913.$ From here we can resort to whatever friendly neighborhood root solving technique floats our Xiddlerian boats to deduce that the next alignments of Planets $A,$ $B$ and $C$ will occur in $\hat{t} = 7.76678$ years. For instance, the secant method gets with initial $a_{-1} = 5, a_0 = 10$ and $$a_n = \frac{a_{n-2}\tilde{S}(a_{n-1}) - a_{n-1} \tilde{S}(a_{n-2})}{\tilde{S}(a_{n-1}) - \tilde{S}(a_{n-2})}$$ with stopping criterion $|\tilde{S}(a_n)| \lt 10^{-6}$ solution quoted above within $6$ steps. The figure below shows the alignment of the planets with respect to their shared sun at the center of their concentric orbits.