Is different sizes items can fit in a box I have checked most of the question in math stack exchange but didn't get the exact answer which I am looking for. I need a solution to a problem
Let say I have 3 items of different size 


*

*12m(l)*4m(w)*3m(h) = 144 volume

*3m(l)*2m(w)*1m(h) = 6 volume

*4m(l)*2m(w)*1m(h) = 8 volume
Large size box: 15m(l)*8m(w)*6m(h) = 720 volume
So want to clarify, to check whether I am able to put these 3 items in a box do I have to sum the volume of items and check whether volume of item is less then volume of box if it's less then I am able to put that item in a box else if volume of a box is less then items volume then I am not able to put those items in a box?
if(items(v) > box(v))
items can not be fit in a box
else
items fit in a box

 A: In one direction the volume test is easy.
If the total volume of the items is greater than the volume of the box then they will not fit inside the box.
If you are allowed to rotate the items then it is irrelevant which dimension is width, which is length, and which is height, so you might as well not bother to label the dimensions. You can just write $12\times 4\times 3,$ for example.
If you are allowed to plastically deform the items into any shape you want, then measuring the volume is enough. But if you must fit the items in the box simultaneously without changing any of their shapes, you must work harder.
For example, consider a $13\times 13\times 13$ box and two
$10\times 10\times 10$ items.
The volume of the box is $2197.$
The combined volume of the items is $2000.$
Now describe where you will put the two items in the box so that both fit.
Another example: take a box of dimensions $15\times 8\times 6$
and take items of dimensions $4\times 15\times 3$ (volume $180$)
and $8\times 8\times 6$ (volume $384$).
The total volume of items to put in the box is $564$, which is less than $720.$
Good. Each item fits in the box (one needs to rotate, but it fits). Good.
But the items do not both fit in the box at the same time
without occupying some of each others' space.
IIRC, this is one of the $NP$-hard problems, which implies there is no quick and easy solution for the general case.
In your particular problem, however, you should easily be able to describe exactly how to arrange the items in the box so that they all fit in the box simultaneously.
Then you will know that you have the right answer.
"Exactly how to arrange the items" could mean that you define the eight corners of the box in Cartesian coordinates, then for each item you define its eight corners in Cartesian coordinates. Then you can check whether any part of any time is outside the box and whether any part of any item is "inside" another item.
Alternatively, you could draw diagrams showing how the items fit.
Your example is easy because you can put all three items on the bottom of the box, so a simple view from above is enough to show that they fit.
A: I have a solution to this problem but I am not sure, If I am wrong then someone comments me out.
First, we can check the lwh of each item and compare it with box lwh. If each of them is lower then box lwh then we can say item can be fit in a box. This rule will apply to each item. 
12m(l)*4m(w)*3m(h) < 15m(l)*8m(w)*6m(h)
3m(l)*2m(w)*1m(h) < 15m(l)*8m(w)*6m(h)
4m(l)*2m(w)*1m(h) < 15m(l)*8m(w)*6m(h)
Second, to check whether all those items can be fit in a large box we will compare the volume of those items with the volume of a large box. If the volume of items is less than then the volume of a box then the items can be fit in a box.
158 volume of items < 720 volume of a box
Confirm both cases, but that will not give you the ability to place the box in any direction. what happens if one of the items has 4m(l)*15m(w)*3m(h) but you can still place this item. 
For this case, I can re-check the lwh of the item with the box lwh by changing its position from vertical to horizontal or from horizontal to vertical.
