Let $$\vec \alpha = 3\hat i + \hat j$$   and $$\vec \beta = 2\hat i - \hat j + 3\hat k.$$
If $$\vec \beta = {{\vec \beta }_1} - {{\vec \beta }_2},$$   where $${{\vec \beta }_1}$$ is parallel to $${\vec \alpha }$$ and $${{\vec \beta }_2}$$ is perpendicular to $${\vec \alpha },$$ then $${{\vec \beta }_1} \times {{\vec \beta }_2}$$   is equal to :

Solution :
$$\vec \beta = {{\vec \beta }_1} - {{\vec \beta }_2}.....(1)$$
Since, $${{\vec \beta }_2}$$ is perpendicular to $${\vec \alpha }.$$
$$\therefore {{\vec \beta }_2}.\vec \alpha = 0$$
Since, $${{\vec \beta }_1}$$ is parallel to $${\vec a}.$$
then $${{\vec \beta }_1} = \lambda \vec \alpha \,\,\left( {{\text{say}}} \right)$$
$$\eqalign{ & \vec a.\vec \beta = \vec a.{{\vec \beta }_1} - \vec \alpha .{{\vec \beta }_2} \cr & \Rightarrow 5 = \lambda {\alpha ^2} \cr & \Rightarrow 5 = 1 \times 10 \cr & \Rightarrow \lambda = \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \left| {\vec \alpha } \right| = \sqrt {10} } \right) \cr & \therefore {{\vec \beta }_1} = \frac{{\vec \alpha }}{2} \cr} $$
Cross product with $${{\vec \beta }_1}$$ in equation (1)
\[\begin{array}{l} \Rightarrow \vec \beta \times {{\vec \beta }_1} = - {{\vec \beta }_2} \times {{\vec \beta }_1}\\ \Rightarrow \vec \beta \times {{\vec \beta }_1} = {{\vec \beta }_1} \times {{\vec \beta }_2}\\ \Rightarrow {{\vec \beta }_1} \times {{\vec \beta }_2} = \frac{{ - {{\vec \beta }_1} \times \vec \alpha }}{2}\\ \Rightarrow {{\vec \beta }_1} \times {{\vec \beta }_2} = \frac{1}{2}\left| \begin{array}{l} \hat i\,\,\,\,\,j\,\,\,\,\,k\\ 2\,\,\, - 1\,\,\,\,3\\ 3\,\,\,\,\,1\,\,\,\,\,0 \end{array} \right|\\ \Rightarrow {{\vec \beta }_1} \times {{\vec \beta }_2} = \frac{1}{2}\left[ { - 3\hat i - \hat j\left( { - 9} \right) + \hat k\left( 5 \right)} \right] \end{array}\]