Ok some start along the way.
First if we order $x_1,x_2,x_3$ as columns we can write the equation system as a matrix:
$$\left[\begin{array}{rrr|r}1&1&0&0\\\0&a&1&0\\\2&1&a&0\end{array}\right]$$
where the first row means $x_1+x_2 = 0$ and so on...
If you replace row 3 with 2 times row 1 minus row 3
$$\left[\begin{array}{rrr|r}1&1&0&0\\\0&a&1&0\\\0&-1&a&0\end{array}\right]$$
Now the tricky part is the lower right $2\times 2$ block, but you find that $a = \pm i$ is special since (i,1) is parallell to $(-1,i)$ which you can realize by multiplying one by $-i$ or the other by $i$.
So unless $a = \pm i$ we can be sure that the solution can only be the 0 vector because for any non-degenerate homogenous equation system only has the zero vectors as solution.
But if $a = \pm i$ we would have to solve it by paramterization, let us assume $x_1 = t$, then by first equation $x_2 = -t$ and then we can pick any of eq 2 or 3 to express $x_3$ in $t$.