find $a,b,c$ such that $ \big| a + b\sqrt{2}+ c\sqrt{3} \big| <10^{-3} $ i know that linear combinations of $1$, $\sqrt{2}$, and $\sqrt{3}$ are dense in the real number line.
$$    \overline{ \mathbb{Z}[\sqrt{2},\sqrt{3}] } = \mathbb{R}    $$
how quickly to these numbers converge to the entire number line. what are the smallest $a,b,c$ we can find so that 
$$   \big| a + b\sqrt{2}+ c\sqrt{3}  \big|  <10^{-3}  \text{  or }10^{-6} $$
by  the pigeonhole principles we can always find $-N<a,b,c<N  $ such that 
$$   \big| a + b\sqrt{2}+ c\sqrt{3}  \big|  <N^{-2}   $$

Here are some lower bounds for the best you can do. 
If $\alpha=a+b\sqrt{2}+c\sqrt{3}$ and $|\alpha|<1$ with $a,b,c$ integers, and $abc\neq 0$, then $$|\alpha|>\frac{1}{(2|b|\sqrt{2}+1)(2|c|\sqrt{3}+1)(2|a|+1)}\tag{1}$$
I get this lower bound since: $\alpha(\alpha-2b\sqrt{2})(\alpha-2c\sqrt{3})(\alpha-2a)$ must be an integer.
If $a=0$, the same trick has that if $|\alpha|<1$ then $|\alpha|>\frac{1}{2|c|\sqrt{3}+1}$.
If $b=0$ or $c=0$, the best you can do is with the continued fractions of $\sqrt{3}$ and $\sqrt{2}$, respectively.
You get (1) in general, for all $a,b,c$.
You tend to get some "good" values when:
$$N(a,b,c)=|\alpha(\alpha-2b\sqrt{2})(\alpha-2c\sqrt{3})(\alpha-2a)|=|a^4+4b^4+9c^4-(4a^2b^2+6a^2c^2+12c^2b^2)|$$
is a small integer. 
For example, $N(14,5,4)=4$ and $14-5\sqrt{2}-4\sqrt{3}$ is a good approximation.
This is because if $|\alpha|<1$ then:
$$\frac{N(a,b,c)}{(2|a|+1)(2|b|\sqrt{2}+1)(2|c|\sqrt{3}+1)}<|\alpha|<\frac{N(a,b,c)}{(2|a|-1)(2|b|\sqrt{2}-1)(2|c|\sqrt{3}-1)}$$
