Given three languages L1 L2 L3 that do not intirsect could one be TR and the other TD and the third neither where $L_{1} \cup L_{2} \cup L_{3} = \sum^{*}$ and 

$L_{1} \cap L_{2} = \emptyset$ and $L_{2} \cap L_{3} = \emptyset$ and $L_{1} \cap L_{3} = \emptyset$

is it possible that $L_{1}$ is decidable, $L_{2}$ is recognizable but not decidable

and $L_{3}$ is not recognizable?

If so please give an example, if not why?

By recognizable I mean Turing Recognizable.

To make things a little easier (I think!) I have reached a conclusion that it is very much possible, I only can't think of an example of such 3 languages.
Is is indeed possible.

Consider the halting problem set $$H = \big\{ \big(\langle M \rangle, w\big): M \big( w \big) \space halts \big\}$$

and its complement $$\overline{H} = \big\{  \big(\langle M \rangle, w\big): M \big( w \big) \space doesn't \space halt \big\}$$

We know that H is recognizable and not decidable. That means that $$\overline{H}$$ is not recognizable because, otherwise, H would be decidable.

Now take $$L_1 = \emptyset \space \big( decidable \big) $$ $$L_2 = H \space \big( recognizable \space and \space not \space decidable \big) $$ $$L_3 = \overline{H} \space \big( not \space recognizable \big)$$

We have that $$L_1 \cup L_2 \cup L_3 = \Sigma^*$$ and
$$L_1 \cap L_2 = L_2 \cap L_3 = L_1 \cap L_3 = \emptyset$$