Given,

$\overrightarrow{a}$ be a non-zero vector parallel to the line of intersection of the two planes described by $\hat{i}+\hat{j},\hat{i}+\hat{k}$, 

So finding the normal vector to the above plane, we get,

${\overrightarrow{n}}_1=\left|\begin{array}{lll}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 1& 0& 1\end{array}\right|=\hat{i}-\hat{j}-\hat{k}$

And normal vector to planes $\hat{i}-\hat{j},\hat{j}-\hat{k}$ will be,

${\overrightarrow{n}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 0\\ 0& 1& -1\end{array}\right|=\hat{i}+\hat{j}+\hat{k}$

Now $\overrightarrow{a}=\lambda \left({\overrightarrow{n}}_1\times {\overrightarrow{n}}_2\right)$

So, direction ratio of $\overrightarrow{a}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& -1\\ 1& 1& 1\end{array}\right|$$=-2\hat{j}+2\hat{k}$

Hence, direction ratio of $\overrightarrow{a}=\left(0,-2,2\right)=\left(0,-1,1\right)$

Now given, $\overrightarrow{b}=2\hat{i}-2\hat{j}+\hat{k}$

And $\overrightarrow{a}=\lambda (-\hat{j}+\hat{k})$

So, $\overrightarrow{a}·\overrightarrow{b}=6=\lambda (2+1)$

$\Rightarrow \lambda =2$

$\therefore \overrightarrow{a}=-2\hat{j}+2\hat{k}$

Now we know that, $\overrightarrow{a}·\overrightarrow{b}=\left|\overline{a}\right|\left|b\right|\cos \theta$

$\Rightarrow 6=2\sqrt{2}\times 3\cos \theta$

$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$

$\Rightarrow \theta =\frac{\pi }{4}$

Now finding, $\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin \theta =2\sqrt{2}\times 3\times \frac{1}{\sqrt{2}}=6$

Hence, ordered pair $\left(\theta ,\left|\overrightarrow{a}\times \overrightarrow{b}\right|\right)=\left(\frac{\mathrm{\pi }}{4},6\right)$