Cho điểm A(1;-2;1) B(0;2;-1) C(2;-3;1) điểm M thỏa mãn T=MA^2-MB^2+MC^2 nhỏ nhất

Giả sử \(M\left( {x;y;z} \right) \Rightarrow \left\{ \begin{array}{l} \overrightarrow {AM} = \left( {x - 1;y + 2;z - 1} \right)\\ \overrightarrow {BM} = \left( {x;y - 2;z + 1} \right)\\ \overrightarrow {CM} = \left( {x - 2;y + 3;z - 1} \right) \end{array} \right. \Rightarrow \left\{ \begin{array}{l} A{M^2} = {\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^2} + {\left( {z - 1} \right)^2}\\ B{M^2} = {x^2} + {\left( {y - 2} \right)^2} + {\left( {z + 1} \right)^2}\\ C{M^2} = {\left( {x - 2} \right)^2} + {\left( {y + 3} \right)^2} + {\left( {z - 1} \right)^2} \end{array} \right.\)

\(\Rightarrow T = \left[ {{{\left( {x - 1} \right)}^2} + {{\left( {y + 2} \right)}^2} + {{\left( {z - 1} \right)}^2}} \right] - \left[ {{x^2} + {{\left( {y - 2} \right)}^2} + {{\left( {z + 1} \right)}^2}} \right] + \left[ {{{\left( {x - 2} \right)}^2} + {{\left( {y + 3} \right)}^2} + {{\left( {z - 1} \right)}^2}} \right]\)

\(= \left[ {{{\left( {x - 1} \right)}^2} - {x^2} + {{\left( {x - 2} \right)}^2}} \right] + \left[ {{{\left( {y + 2} \right)}^2} - {{\left( {y - 2} \right)}^2} + {{\left( {y + 3} \right)}^2}} \right] + \left[ {{{\left( {z - 1} \right)}^2} - {{\left( {z + 1} \right)}^2} + {{\left( {z - 1} \right)}^2}} \right]\)\(= \left( {{x^2} - 6x + 5} \right) + \left( {{y^2} + 14y + 17} \right) + \left( {{z^2} - 6z + 1} \right)\)

\(= {\left( {x - 3} \right)^2} - 4 + {\left( {y + 7} \right)^2} - 32 + {\left( {z - 3} \right)^2} - 8 \ge - 4 - 32 - 8 = - 44.\)

Dấu "=" xảy ra khi \(x = 3,{\rm{ }}y = - 7,{\rm{ }}z = 3.\)

Khi đó \(M\left( {3; - 7;3} \right) \Rightarrow P = x_M^2 + 2y_M^2 + 3z_M^2 = 134.\)