If two vectors are parallel then they are proportional.

So, let $\overrightarrow{{\beta }_1}=k\overrightarrow{\alpha }$

Given $\overrightarrow{\beta }=\overrightarrow{{\beta }_1}-\overrightarrow{{\beta }_2}$

$\Rightarrow \overrightarrow{\alpha }\cdot \overrightarrow{\beta }=\overrightarrow{\alpha }\cdot \overrightarrow{{\beta }_1}-\overrightarrow{\alpha }\cdot \overrightarrow{{\beta }_2} ...\left(i\right)$

Now, $\overrightarrow{\alpha }·\overrightarrow{\beta }=\left(3\hat{i}+\hat{j}\right)·\left(2\hat{i}-\hat{j}+3\hat{k}\right)$

$=6-1=5$

And, $\overrightarrow{\alpha }$ is perpendicular to $\overrightarrow{{\beta }_1},$ hence $\overrightarrow{\alpha }\cdot \overrightarrow{{\beta }_1}=0$

And, $\overrightarrow{\alpha }\cdot \overrightarrow{{\beta }_2}=\overrightarrow{\alpha }\cdot \left(k\overrightarrow{\alpha }\right)=k{\left(\overrightarrow{\alpha }\right)}^2=k{\left|\overrightarrow{\alpha }\right|}^2$

Also, $\left|\overrightarrow{\alpha }\right|=\sqrt{3^2+1^2}=\sqrt{10}$

Put, all these values in the equation $\left(i\right),$ to get $5=k\times 10$

$\therefore k=\frac{1}{2}$

$\therefore \overrightarrow{{\beta }_1}=\frac{1}{2}(3\widehat{i}+\widehat{j})$

$=\frac{3}{2}\widehat{i}+\frac{1}{2}\widehat{j}$

Now $\overrightarrow{{\beta }_2}=\overrightarrow{{\beta }_1}-\overrightarrow{\beta }=-\frac{1}{2}\widehat{i}+\frac{3}{2}\widehat{j}-3\widehat{k}$

$\therefore \overrightarrow{{\beta }_1}\times \overrightarrow{{\beta }_2}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ \frac{3}{2}& \frac{1}{2}& 0\\ -\frac{1}{2}& \frac{3}{2}& -3\end{array}\right|$

$=\widehat{i}\left(-\frac{3}{2}-0\right)-\widehat{j}\left(-\frac{9}{2}-0\right)+\widehat{k}\left(\frac{9}{4}+\frac{1}{4}\right)$

$=-\frac{3}{2}\widehat{i}+\frac{9}{2}\widehat{j}+\frac{5}{2}\widehat{k}$

$=\frac{1}{2}(-3\widehat{i}+9\widehat{j}+5\widehat{k}).$