Let $u=(a,b,c)$ transpose with a and c not equal to 0. Find real numbers x and y such that the vector $v = (b, -a, x)^T$ is orthogonal to $u$, and the vector $w = (a,b,y)^T$ is orthogonal to both $u$ and $w$.
My steps:
$$(a,b,c)^T\cdot (b,-a,x)^T = 0$$
So, $ab - ab + cx = 0$, $cx = 0$ and knowing that c is not equal to zero, $x = 0$
Then,
$$(a,b,c)^T\cdot (a,b,y)^T = 0$$
So, $a^2+b^2+cy=0$ and knowing that a is not equal to zero means that $(a^2+b^2)>0$, therefore either $c>0$ and $y<0$ or $c<0$ and $y>0$
And finally,
$$(b,-a,x)^T \cdot (a,b,y)^T = 0$$
So, $ab - ab + xy = 0$, so $xy = 0$ which is true regardless of $y$, because $x = 0$
So essentially, I get 2 scenarios:
$c>0 \implies x=0$ and $y<0$
$c<0 \implies x=0$ and $y>0$
However, I have a feeling this is not precise enough for the answer.
You are almost there, but instead of complicating matters with inequalities, you can simply express $y$, the solution is unque:
$$a^2+b^2+cy=0\rightarrow y=-\frac{a^2+b^2}{c}$$
And the last step is still true, so this $y$ also ensures orthogonality with $v$.