Solving for a three dimensional vector. Let
$a = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix}$ and $b = \begin{pmatrix} -6 \\ 4 \\ -3 \end{pmatrix}$  
There exists two nonzero three-dimensional vectors
${v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$
that are orthogonal to both ${a}$ and ${b}$, such that its entries $x$, $y$, and $z$ are integers, that satisfy $\gcd(x,y,z) = 1$. Find either vector.
I have tried to write out the equations, but there aren't enough equations and too many variables.  I still can't figure out how to use $\gcd(x,y,z) = 1$. Any help is appreciated.
 A: Hint: The orthogonal complement of $span(a,b)$ has dimension $1$. To find this set, use Gram-Schmidt to orthogonalize first $a$ and $b$ and then find a vector $c$ that is orthogonal to the plane $span(a,b)$. Now solve the question regarding integer components with $gcd=1$.
A: \begin{align}
2x + 5y - z = 0 \rightarrow x + \frac{5}{2}y - \frac{1}{2}z = 0\\
-6x + 4y - 3z = 0 \rightarrow x - \frac{2}{3}y + \frac{1}{2}z = 0\\
\rightarrow \\
\left(\frac{5}{2} + \frac{2}{3}\right)y - z = 0
\end{align}
This gives that $z = \frac{19}{6}y$ which gives that $x = -\frac{5}{2}y + \frac{1}{2}\frac{19}{6}y = -\frac{11}{12}y$.  This means that $y$ must be divisible by $12$ (to produce an integer).  this gives that $x = -11$, $y = 12$, and $z = 38$.  The gcd of these is $1$ because one of the values is a prime and none of the other values have $11$ as a factor (there is only one value divisible by $3$ and only one divisible by $19$ but there are two values divisible by $2$).
...the other one is just the negation of this one: $x = 11$, $y = -12$, $z = -38$.
