Trong không gian $\left( Oxyz \right)$, cho $A\left( -1;2;0 \right)$, $B\left( 3,-1,0 \right)$. Điểm $C\left( a;b;0 \right)\left( b>0 \right)$ sao...

Ta có: $\overrightarrow{AB}=\left( 4;-3;0 \right);\overrightarrow{BC}=\left( a+1;b-2;0 \right)$ ; $\left[ \overrightarrow{AB};\overrightarrow{AC} \right]=\left( 0;0;4b+3a-5 \right)$.
Vì $\Delta ABC$ cân tại $B\Leftrightarrow A{{B}^{2}}=B{{C}^{2}}\Leftrightarrow {{\left( a-3 \right)}^{2}}+{{\left( b+1 \right)}^{2}}=25\left( 1 \right)$.
Mặt khác: ${{S}_{\Delta ABC}}=\dfrac{25}{2}\Leftrightarrow \dfrac{1}{2}\left[ \overrightarrow{AB};\overrightarrow{AC} \right]=\dfrac{25}{2}$
$\Leftrightarrow \dfrac{1}{2}\left| 3a+4b-5 \right|=\dfrac{25}{2}\Leftrightarrow \left| 3a+4b-5 \right|=25$
$\Leftrightarrow \left[ \begin{aligned}
& 3a+4b-5=25 \\
& 3a+4b-5=-25 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& 3a+4b=30 \\
& 3a+4b=-20 \\
\end{aligned} \right.$.
TH1: $3a+4b=30\Leftrightarrow a=\dfrac{30-4b}{3}$. Thay vào $\left( 1 \right)$ ta được ${{\left( \dfrac{30-4b}{3}-3 \right)}^{2}}+{{\left( b+1 \right)}^{2}}=25$
$\begin{aligned}
& \Leftrightarrow {{\left( 21-4b \right)}^{2}}+9{{\left( b+1 \right)}^{2}}=225 \\
& \Leftrightarrow 25{{b}^{2}}-150b+295=0 \\
& \Leftrightarrow {{b}^{2}}-6b+9=0 \\
& \Leftrightarrow b=3\Rightarrow a=6 \\
\end{aligned}$
Vậy $T={{3}^{2}}+{{6}^{2}}=45$.
TH2: $3a+4b=-20\Leftrightarrow a=\dfrac{-20-4b}{3}$
Thay vào $\left( 1 \right)$ ta được ${{\left( \dfrac{-20-4b}{3}-3 \right)}^{2}}+{{\left( b+1 \right)}^{2}}=25$
$\Leftrightarrow {{\left( 29+4b \right)}^{2}}+9{{\left( b+1 \right)}^{2}}=225$
$\Leftrightarrow 25{{b}^{2}}+241b+625=0$ (vô nghiệm).
Vậy $T=45$.