Goi P(a,b)$\rightarrow$$\overrightarrow{PA}$=(3-a,-b,-1)
$\overrightarrow{PB}$=(1-a,3-b,-2)
$\overrightarrow{PC}$=(3-a,-4-b,1)
$\left| {PA+PB+PC} \right|$min$\Leftrightarrow (7-3a)^{2}$+$(1+3b)^{2}$+4$\geq $4 vs moi a,b
dấu '=' xảy ra $\Leftrightarrow$7-3a=0 hoặc 1+3b=0
$\rightarrow $P(7/3;-1/3;0)