How can we get 1st index elements of each array in different new array

for getting all the three arrays as desired/required

a!localVariables(
  local!array:{
    {
      list: {2,13,65}
    },
    {
      list: {7,15,45}
    },
    {
      list: {14,3,56}
    },
    {
      list: {1,2,3}
    }

  },
  a!forEach(
    items:enumerate(3)+1,  /* three is the no. arrays you have to make/ output arrays */
    expression: a!localVariables(
      local!temp:fv!item,
      a!forEach(
    items: local!array,
    expression: fv!item.list[local!temp]
  )))
)

What is the point of copy-pasting the entire code but with increased complexity - one more a!forEach()? That just adds more executing time. 

Sir, as he said in the question that he wants n(3) arrays accordingly .. so I thought what if the arrays are {1,2,3,4,.....,n} ... So i've edited your code with one extra loop in which he can put the number of required arrays for output. 
  eg - [1,2,3] , [2,4,6], [4,8,9]  should be [1,2,4],[2,4,8],[3,6,9] 
     
so i've added an extra loop for getting all the arrays as - if you put 2 in the first loop than you can get only [1,2,4],[2,4,6]. and same will work for the n times with a index function also.
please correct me if I'm wrong

Sir, as he said in the question that he wants n(3) arrays accordingly .. so I thought what if the arrays are {1,2,3,4,.....,n} ... So i've edited your code with one extra loop in which he can put the number of required arrays for output. 
  eg - [1,2,3] , [2,4,6], [4,8,9]  should be [1,2,4],[2,4,8],[3,6,9] 
     
so i've added an extra loop for getting all the arrays as - if you put 2 in the first loop than you can get only [1,2,4],[2,4,6]. and same will work for the n times with a index function also.
please correct me if I'm wrong

Ah, okay! Then it makes sense.