Question 1069910
We see that the numbers {{{0}}} , {{{1}}} , {{{2}}} , and {{{5}}} appear three times each.
A string of twelve whole numbers separated by {{{"+"}}} and {{{"-"}}} signs
can be considered a sum of twelve integer numbers, some positive, some negative, some zero.
We can change the order of terms in that sum (commutative property),
and we can groups numbers together (associative property)
as we wish to make it more convenient for our calculations.

You can safely place {{{"-"}}} signs in front of the zeros,
because it does not matter if {{{0}}} is "added" or "subtracted".
  
To minimize the number of {{{"+"}}} signs,
you would want then to be placed before the greatest whole numbers,
while placing {{{"-"}}} signs in front of whole smaller numbers.
We know that {{{2+1+1+1=5}}} <--> {{{0=5-(2+1+1+1)=5+(-2)+(-1)+(-1)+(-1)}}} .
We can place a {{{"+"}}} sign in front of a {{{5}}} ,
and place {{{"-"}}} signs in front of every {{{1}}} and a {{{2}}} .
Having {{{red(1)}}} {{{"+ 5"}}} counteracting one {{{-2}}} and three {{{-1}}} terms
gives you a sum of {{{0}}} for those terms.
After doing all that, there are only 3 asterisks left to be replaced,
one in front of {{{2}}} , and two in front of {{{5}}} .
The {{{2}}} can take a {{{"-"}}} sign to neutralize the {{{2}}} at the start, which is positive but does not need a sign.
Then {{{red(1)}}}  {{{5}}} can take a {{{"+"}}} sign,
and the other {{{5}}} can get a {{{"-"}}} sign with the total sum being zero.
It does not matter which {{{5}}} gets the {{{"-"}}} sign.
{{{2-0-1+5-2-0-1-5-2-0-1+5 = 0}}} ,
{{{2-0-1-5-2-0-1+5-2-0-1+5 = 0}}} , and
{{{2-0-1+5-2-0-1+5-2-0-1-5 = 0}}} are all true,
and each one has only {{{red(1)+red(1)=highlight(2)}}} {{{"+"}}} signs.