$a^2+b^2+c^2$

There are many solutions, but only one possible with all integer values. So let's search for integers solutions.

By hypotesys we can assume $b = a + r$ and $c = a + 2r$

We have then

$3a^2 + 6ar + 5r^2 = 83$

Let $m$ be the minimum value between $a$ and $r$ . We have

$11 m^2 \leq 83$

So $m^2 \leq {83 \over 11}$ and in the end

$m \leq 2$ .

So we know that at least one between $a$ and $r$ must be $1$ or $2$ .

If we try $a = 1$ and $a = 2$ it leads $r$ to assume non integer values. Also taking $r = 1$ leads non integer solutions for $a$ . Taking the only other option $r = 2$ gives the following equation for $a$

$3a^2 +12a +20 = 83$

that is

$a^2 + 4a +4 = 25$

which gives the only positive solution $a = 3$ .

So we have

$a=3, b = a+r =5, c = a + 2r = 7$

and the sum is $15$ .

${ 3 }^{ 2 }+{ 5 }^{ 2 }+{ 7 }^{ 2 }=83\\ 3+5+7=15$