$\begin{align*}
&求直線L與M的交點P：\\
&\begin{cases} x+3y=0 \\ 2x-3y+9=0 \end{cases} \implies P(-3,1)。\\
\\
&由\overrightarrow{PB_1}=\frac{1}{2}\overrightarrow{PB_3}，設B_3(x,y)，已知B_1(3,2)，得：\\
&(3,2)-(-3,1)=\frac{1}{2}\left[(x,y)-(-3,1)\right] \\
&\implies (6,1)=\frac{1}{2}(x+3,y-1) \implies (x,y)=(3,5)。\\
\\
&故B_3坐標為(3,5)。
\end{align*}$

https://www.ceec.edu.tw/files/file_pool/1/0n045357541158913049/04-112%e5%ad%b8%e6%b8%ac%e6%95%b8%e5%ad%b8b%e9%81%b8%e6%93%87%28%e5%a1%ab%29%e9%a1%8c%e7%ad%94%e6%a1%88.pdf