Ok well if we have the sum of these three things, we can find a cubic equation which can be solved.

Suppose our answers are a,b and c.

(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc)

So (ab+ac+bc) = ((a+b+c)^2 - (a^2+b^2+c^2))/2

Now (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc

So our equation is

x^3 - (a+b+c)x^2 + ((a+b+c)^2-(a^2+b^2+c^2))/2 - abc = 0

In the case above we have

x^3 - 11x^2 + (121-53)/2x - 24 = x^3 - 11x^2 + 34x - 24

We can solve this using analytical methods to get roots = 1,4,6

x+y+z=11

xyz=24

x^2+y^2+z^2 = 53

(1,4,6) is the answer.

I just tried a bunch of values. I am not sure of how to solve a problem like this in general. I will think about it, and see if there is a method.

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