Question 1154151
Let the 3 odd integers be x - 2, x and x + 2
Square of the 1st = {{{x^2 - 4x + 4}}}
Square of the 2nd = {{{x^2}}}
Square of the 3rd = {{{x^2 + 4x + 4}}}
Adding the 3 equations above, and equating to the given sum, we get
{{{3x^2 + 8 = 371}}}

{{{3x^2 = 371 - 8 = 363}}} 
{{{x^2 = 363/3 = 121}}}

So {{{x = sqrt(121) = 11}}}
since it is given that x is positive, it cannot be -11

So the 3 numbers are {{{9}}}, {{{11}}} and {{{13}}}.

Check by adding the 3 squares
{{{9^2 + 11^2 + 13^2 = 81 + 121 + 169 = 371}}}. Checked!

<b>Ans: 9, 11, 13</b>