Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$. Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$.
By distinct it means that $(1, 0, 0)$ is a solution, but $(0, \pm 1, 0)$ counts as the same solution.
I can only seem to find two; namely $(1, 0, 0)$ and $( \frac{-1}{3}, \frac{2}{3}, \frac{2}{3})$. Is there a method to finding a third or is it still just trial and error?

There are infinitely many. The complete rational solution to
$$a^2+b^2+c^2=1$$
is given by
$$\left(\frac{p^2-q^2-r^2}s\right)^k+\left(\frac{2pq}s\right)^k+\left(\frac{2pr}s\right)^k=1\tag1$$
where $s=p^2+q^2+r^2$ and $k=2$. But eq $(1)$ is also satisfied for $k=1$ if 
$$p=\frac{q^2+r^2}{q+r}$$
For example, if $q=1,\,r=5$, then,
$$\big({-}\tfrac{5}{31}\big)^k+\big(\tfrac{6}{31}\big)^k+\big(\tfrac{30}{31}\big)^k=1$$
for $k=1,2$, and so on.
