Squares In Between

$(a + b + c)^3 = a^3 + b^3 + c^3 + (3ab^2 + 3ac^2 + 3ba^2 + 3bc^2 + 3ca^2 + 3cb^2 +6abc)$

$(a + b + c)^3 - (a^3 + b^3 + c^3) = 3(ab^2 + ac^2 + ba^2 + bc^2 + ca^2 + cb^2 + 2abc) = 3(a + b)(b + c)(c + a)$

From the system above, $(a + b + c)^3 - (a^3 + b^3 + c^3) = 11^3 - 785 = 546 = 3(a + b)(b + c)(c + a)$

Hence, $(a + b)(b + c)(c + a) = 182 = 2\times 7\times 13$ .

Since $a, b, c$ are integers, each sum of the two variables will equal to each of the three primes presented. Covering the cyclic solutions, we can set up the new system of equations as followed:

$a + b = 2$ $b + c = 7$ $c + a = 13$

Then $c -a = 5$ , and $2c = 18; c = \boxed{9}$ .

Hence, $a = \boxed{4}$ , and $b = \boxed{-2}$ .

Thus, $a^2 + b^2 + c^2 = 4^2 + (-2)^2 + 9^2 = \boxed{101}$ .