The figure from below shows a cube of side $a$ where $C$ and $D$ are midpoints of the edges. Find a unit vector paralell to vector $\vec{x}$ such that $\overrightarrow{AC}\cdot \vec{x}=0$ and $\overrightarrow{BD}\cdot\vec{x}=0$

The alternatives are as follows:

$\begin{array}{ll}

1.&\left(\frac{\sqrt{17}}{17}\right)(\hat{j}+4\hat{k})\\

2.&\left(\frac{\sqrt{2}}{2}\right)(-\hat{j}+\hat{k})\\

3.&\left(\frac{\sqrt{5}}{5}\right)(\hat{j}+2\hat{k})\\

4.&\left(\frac{\sqrt{10}}{10}\right)(3\hat{j}+\hat{k})\\

5.&(\frac{\sqrt{26}}{26})(\hat{j}+5\hat{k})\\

\end{array}$

This problem has left me to go in circles. The only thing which I was able to establish in this cube was that the vectors

$\overrightarrow{BD}=a(\left \langle +1,+1, -\frac{1}{2} \right \rangle)$

$\overrightarrow{AC}=a(\left \langle -1,+1, -\frac{1}{2} \right \rangle)$

However this is the part where I got stuck. How exactly can I find the vector which is requested?. Help! please.