Let, $\overrightarrow{a}$ makes an angle $\theta$ with $Z$-axis and $\overrightarrow{b}$ makes an angle $\varphi$ from $X$-axis.

Apply ${\cos }^{2}60°+{\cos }^{2}60°+{\cos }^2\theta =1 \& {\cos }^2\varphi +{\cos }^{2}45°+{\cos }^{2}45°=1$

$\Rightarrow \cos \theta =\frac{\pm 1}{\sqrt{2}} \& \cos \varphi =0$

So, $\overrightarrow{a}=\left(2\cos {60}^{\circ }\right)\hat{i}+\left(2\cos {60}^{\circ }\right)\hat{j}+\left(2\cos \theta \right)\hat{k}$ and $\overrightarrow{b}=\left(\sqrt{2}\cos \varphi \right)\hat{i}+\left(\sqrt{2}\cos {45}^{\circ }\right)\hat{j}+\left(\sqrt{2}\cos {45}^{\circ }\right)\hat{k}$

$\Rightarrow \overrightarrow{a}=\hat{i}+\hat{j}\pm \sqrt{2}\hat{k}$ and $\overrightarrow{b}=\hat{j}+\hat{k}$

Now, $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& \pm \sqrt{2}\\ 0& 1& 1\end{array}\right|$

$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\left(1\pm \sqrt{2}\right)\hat{i}-\hat{j}+\hat{k}$