giúp em với

Chọn điểm $\displaystyle I$ sao cho: $\displaystyle \overrightarrow{IA} +2\overrightarrow{IB} +\overrightarrow{IC} =\vec{0}$ (tâm tỉ cự)
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Longrightarrow \ ( 1-x;0-y;5-z) +2( 2-x;1-y;-1-z) +( 3-x;-2-y;1-z) =\vec{0}\\
\Leftrightarrow \begin{cases}
1-x+4-2x+3-x=\vec{0}\\
-y+2-2y-2-y=\vec{0}\\
5-z-2-2z+1-z=\vec{0}
\end{cases} \Leftrightarrow \begin{cases}
x=2\\
y=0\\
z=1
\end{cases} \Longrightarrow I( 2;0;1)
\end{array}$

Ta có: $\displaystyle P=MA^{2} +2MB^{2} +MC^{2} =(\overrightarrow{MI} +\overrightarrow{IA})^{2} +2(\overrightarrow{MI} +\overrightarrow{IB})^{2} +(\overrightarrow{MI} +\overrightarrow{IC})^{2}$
             $\displaystyle  \begin{array}{{>{\displaystyle}l}}
=\overrightarrow{MI}^{2} +2\overrightarrow{MI} .\overrightarrow{IA} +\overrightarrow{IA}^{2} +2\overrightarrow{MI}^{2} +4\overrightarrow{MI} .\overrightarrow{IB} +2\overrightarrow{IB}^{2} +\overrightarrow{MI}^{2} +2\overrightarrow{MI} .\overrightarrow{IC} +\overrightarrow{IC}^{2}\\
=4\overrightarrow{MI}^{2} +2\overrightarrow{MI}(\overrightarrow{IA} +2\overrightarrow{IB} +\overrightarrow{IC}) +\overrightarrow{IA}^{2} +2\overrightarrow{IB}^{2} +\overrightarrow{IB}^{2}\\
=\ 4\overrightarrow{MI}^{2} +\overrightarrow{IA}^{2} +2\overrightarrow{IB}^{2} +\overrightarrow{IB}^{2}
\end{array}$
$\displaystyle \Longrightarrow P_{min} \ khi\ 4\overrightarrow{MI}^{2}$nhỏ nhất hay $\displaystyle MI$ nhỏ nhất $\displaystyle \Longrightarrow \ M\ \equiv \ I$
⟹ $\displaystyle M( 2;0;1) .$
Vậy, $\displaystyle T=2+0+1=3.$ $