We have 3 arithmetic progressions $na_1 + c_1, na_2 + c_2, na_3 + c_3$ with the reals $a_1,a_2,a_3$ linearly independent over the rationals. Is it always true that given $\epsilon > 0$ there are terms $x_1,x_2,x_3$, one from each progression, such that all three are within $\epsilon$ of each other ?
Same problem with any finite number of progressions.
This is actually not true. For a counterexample, take $(a_1,a_2,a_3) = (1,\phantom. \pi,\phantom.\pi/(\pi+1))$ and $(c_1,c_2,c_3) = (0,0,1)$, so our sequences are $\lbrace n_1 \rbrace$, $\lbrace n_2 \pi \rbrace$, and $\lbrace 1 + n_3 \frac\pi{\pi+1}\rbrace$. If $n_1$ is near $n_2 \pi$ then it's also near $(n_1+n_2)\frac\pi{\pi+1}$, and therefore far from $1 + n_3 \frac\pi{\pi+1}$ for any integer $n_3$.
As this example suggests, the correct condition (for any number of arithmetic progressions) is that the reciprocals of the common differences $a_i$ be linearly independent over ${\bf Q}$.
Indeed, given nonzero $a_1,\ldots,a_{k+1} \in {\bf R}$, consider the image of the lattice $L = \oplus_{i=1}^{k+1} {\bf Z} a_i \subset {\bf R}^{k+1}$ under the quotient map $\pi : {\bf R}^{k+1} \rightarrow V := {\bf R}^{k+1} / {\bf R}v$ where $v=(1,1,\ldots,1)$. Then the desired result is true for all $c_1,\ldots,c_{k+1}$ iff $\pi(L)$ is dense in $V$. But $\pi(L)$ is a subgroup of $V$, so its closure is a closed subgroup — and a closed subgroup $G \subseteq V\phantom.$ is properly contained in $V\phantom.$ iff there a nonzero linear functional $\varpi: V \rightarrow {\bf R}$ such that $\varpi(G) \subseteq \bf Z$. Hence $\pi(L)$ fails to be dense in $V\phantom.$ iff there is a nonzero functional $V \rightarrow {\bf R}$ taking $\pi(L)$ to a subset of ${\bf Z}$. But the linear functionals on $V\phantom.$ are simply the functionals on ${\bf R}^{k+1}$ vanishing on $v$; that is, the maps $(x_1,\ldots,x_{k+1}) \mapsto \sum_{i=1}^{k+1} b_i x_i$ for some $b_i\in\bf R$, not all zero, such that $\sum_{i=1}^{k+1} b_i = 0$. Hence our condition is $b_i a_i \in \bf Z$, or equivalently $b_i \in {\bf Z} a_i^{-1}$. There exist such $b_i$ that sum to zero iff the $a_i^{-1}$ are linearly dependent over the rationals, QED.
This follows from the Kronecker's approximation theorem (With the conditions modified. See edit below).
Assume without loss of generality that $a_k$ are positive, that $c_1 \leq c_k$ and $0 \leq c_k < a_k$. Let $n_1=n$ and $n_k=\lfloor na_1/a_k\rfloor$ for $k=1,\ldots,K$. Then $n_1 a_1+c_1-(n_k a_k+c_k)=n a_1 -\lfloor na_1/a_k\rfloor a_k+c_1-c_k=c_1-c_k + $ { $ n a_1/a_k $ }$a_k$
By Kronecker's approximation theorem there exists some $n$ such that $|c_1-c_k + $ {$n a_1/a_k $}$a_k|<\epsilon/2$ for each $k=1,\ldots,K$. The conclusion follows from the triangle inequality.
Edit: As Noam D. Elkies remarked in his answer what we use here is in fact the linear independence over $\mathbb Q$ of the reciprocals of the numbers, $1/a_k$ (Or equivalently in my application of the Kronecker's approximation theorem, the numbers $a_1/a_k$ ), not the numbers $a_k$ themselves. This means that the question as posed is not true, but it is true when the conditions are modified.
For given $(a_1, a_2, a_3)$ and $(c_1, c_2 , c_3)$, the required condition is equivalent to: $(c_2-c_1, c_3-c_2)$ is in the closure $\Gamma $ of the additive subgroup generated by $(a_1,0)$, $(-a_2,a_2)$, $(0,-a_3)$. But the latter need not be dense, even if $(a_1, a_2, a_3)$ are linearly independent over the rationals; for instance $(a_1, a_2, a_3):=(1+\lambda,1,1+1/\lambda)$ with $\lambda$ an irrational non quadratic, is a linear independent triple for which $\Gamma$ is a family of lines $\{(x,y)\, : \, y=\lambda x + n(\lambda+1)\, , n\in\mathbb{N}\}$.
