Solution:

$\begin{array}{l}
\vec{a}=3 \hat{\imath}-\hat{\jmath}
\end{array}$
$\begin{array}{l}
\mathrm{I} \vec{a} \mathrm{I}=\sqrt{3^{2}+1^{2}}=\sqrt{9+1}=\sqrt{10} \\
\hat{a}=\frac{3}{\sqrt{10}} \hat{\imath}-\frac{1}{\sqrt{10}} \hat{\jmath} \\
\vec{b}=2 \hat{\imath}+\hat{\jmath}-3 \hat{k}=\overrightarrow{b_{1}}+\overrightarrow{b_{2}}
\end{array}$
Given $\overrightarrow{b_{1}}$ is parallel to $\vec{a}$
$\overrightarrow{b_{1}}=\mathrm{K}\left(\frac{3}{\sqrt{10}} \hat{\imath}-\frac{1}{\sqrt{10}} \hat{\jmath}\right)—–I$
Also $\overrightarrow{b_{2}}$ is perpendicular to $\vec{a}$
Let $\overrightarrow{b_{2}}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}$
$\begin{array}{l}
\overrightarrow{b_{2}} \cdot \vec{a}=0 \\
3 x-y=0 \\
3 x=y———-II
\end{array}$
According to the question
$2 \hat{\imath}+\hat{\jmath}-3 \hat{k}=\mathrm{K}\left(\frac{3}{\sqrt{10}} \hat{\imath}-\frac{1}{\sqrt{10}} \hat{\jmath}\right)+\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{k}$
Comparing left hand side and right hand side
$\begin{array}{l}
2=\frac{3 K}{\sqrt{10}}+\mathrm{x}———III \\
1=\frac{-K}{\sqrt{10}}+\mathrm{y}-\mathrm——–{IV} \\
-3=\mathrm{Z}
\end{array}$
Solving III and IV and I we get
$\mathrm{K}=\frac{\sqrt{10}}{2}, \mathrm{x}=\frac{1}{2}, \mathrm{y}=\frac{3}{2}, \mathrm{z}=-3$
So
$\begin{array}{l}
\overrightarrow{b_{1}}=\frac{3}{2} \hat{l}-\frac{1}{2} \hat{\jmath} \\
\overrightarrow{b_{2}}=\frac{1}{2} \hat{\imath}+\frac{3}{2} \hat{\jmath}-3 \hat{k}
\end{array}$