Trong không gian \[Oxyz,\] cho ba điểm \[{\rm{A}}\left( {0\,;\,\,0\,;\,\, - 1} \right),\,\,{\rm{B}}\left( { - 1\,;\,\,1\,;\,\,0} \right),{\rm{C}}\left( {1\,;\,\,0\,;\,\,1} \right)\]. Tìm điểm \({\rm{M}}\) sao cho \(3{\rm{M}}{{\rm{A}}^2} + 2{\rm{M}}{{\rm{B}}^2} - {\rm{M}}{{\rm{C}}^2}\) đạt giá trị nhỏ nhất.

Giả sử \[{\rm{M}}\left( {{\rm{x}}\,;\,\,{\rm{y}}\,;\,\,{\rm{z}}} \right) \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\overrightarrow {{\rm{AM}}}  = \left( {{\rm{x}}\,;\,\,{\rm{y}}\,;\,\,{\rm{z}} + 1} \right)}\\{\overrightarrow {{\rm{BM}}}  = \left( {{\rm{x}} + 1\,;\,\,{\rm{y}} - 1\,;\,\,{\rm{z}}} \right)}\\{\overrightarrow {{\rm{CM}}}  = \left( {{\rm{x}} - 1\,;\,\,{\rm{y}}\,;\,\,{\rm{z}} - 1} \right)}\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{\rm{A}}{{\rm{M}}^2} = {{\rm{x}}^2} + {{\rm{y}}^2} + {{\left( {{\rm{z}} + 1} \right)}^2}}\\{{\rm{B}}{{\rm{M}}^2} = {{\left( {{\rm{x}} + 1} \right)}^2} + {{\left( {{\rm{y}} - 1} \right)}^2} + {{\rm{z}}^2}}\\{{\rm{C}}{{\rm{M}}^2} = {{\left( {{\rm{x}} - 1} \right)}^2} + {{\rm{y}}^2} + {{\left( {{\rm{z}} - 1} \right)}^2}}\end{array}} \right.} \right.\]

\( \Rightarrow 3{\rm{M}}{{\rm{A}}^2} + 2{\rm{M}}{{\rm{B}}^2} - {\rm{M}}{{\rm{C}}^2}\)

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

\( = 4{x^2} + 4{y^2} + 4{z^2} + 6x - 4y + 8z + 6 = {\left( {2x + \frac{3}{2}} \right)^2} + {\left( {2y - 1} \right)^2} + {\left( {2z + 2} \right)^2} - \frac{5}{4} \ge  - \frac{5}{4}.\)

Dấu  xảy ra \( \Leftrightarrow x =  - \frac{3}{4},\,\,y = \frac{1}{2},\,\,z =  - 1\), khi đó \(M\left( { - \frac{3}{4};\,\,\frac{1}{2};\,\, - 1} \right).\) Chọn D.