Not so good-looking, seriously!

First, note that $x = 2, y = -1, z = -1$ is a solution with $x+y+z = 0$ . Now suppose $(x,y,z)$ achieves the minimum sum.

First, if $0 < x \le 1$ , then $(x,y,z)$ is not a minimum sum, since $(-x, y, z)$ is also a solution with smaller sum. Similar with $y, z$ . So we may assume $x,y,z \in [-1, 0] \cup (1, 2]$ . This divides them into two intervals, the negatives $[-1, 0]$ (we consider 0 negative for this purpose) and the positives $(1, 2]$ .

Suppose there are two positives, then $x+y+z > 1+1+(-1) > 0$ so it doesn't achieve the minimum sum. Suppose there are no positives, then $x^2 + y^2 + z^2 \le (-1)^2 \cdot 3 < 6$ , so it doesn't satisfy the condition. So there is exactly one positive.

Finally, now we have $x^2 + y^2 + z^2 \le 2^2 + (-1)^2 + (-1)^2 = 6$ . Since equality is achieved, this (and its permutations) is the only solution. So the minimum sum is 0.