The problem is as follows:
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?.
As you have found both $\vec {AC}$ and $\vec {BD}$, whats the problem in finding $\vec x\,$? From the first two equations, it is evident that $\vec x$ is perpendicular to both $\vec {AC}$ and $\vec {BD}$. Hence all you have to do is find the cross product of the two and divide by its magnitude to get the required unit vector. The answer will come out to be option $3$.