Mặt cầu (left( S right)) có tâm (Ileft( { – 1;0;3} right)), bán kính (R = 1).
Gọi (Jleft( {a;b;c} right)) là điểm thỏa mãn (overrightarrow {JA} + 2overrightarrow {JB} = overrightarrow 0 ).
Ta có: (overrightarrow {JA} = left( {3 – a,,,1 – b,,, – 3 – c} right);,,overrightarrow {JB} = left( { – a;,, – 2 – b;,,3 – c} right))
( Rightarrow overrightarrow {JA} + 2overrightarrow {JB} = left( {3 – 3a;,, – 3 – 3b;,,3 – 3c} right) = overrightarrow 0 Leftrightarrow left{ begin{array}{l}a = 1\b = – 1\c = 1end{array} right. Rightarrow Jleft( {1; – 1;;1} right))
Khi đó ta có:
(begin{array}{l}T = M{A^2} + 2M{B^2} = {left( {overrightarrow {MJ} + overrightarrow {JA} } right)^2} + 2{left( {overrightarrow {MJ} + overrightarrow {JB} } right)^2}\T = M{J^2} + 2overrightarrow {MJ} .overrightarrow {JA} + J{A^2} + 2M{J^2} + 4overrightarrow {MJ} .overrightarrow {JB} + 2J{B^2}\T = 3M{J^2} + 2overrightarrow {MJ} underbrace {left( {overrightarrow {JA} + 2overrightarrow {JB} } right)}_{overrightarrow 0 } + underbrace {J{A^2} + 2J{B^2}}_{const}end{array})
Do đó ({T_{max }} Leftrightarrow M{J_{max }}).
Ta có: (overrightarrow {IJ} = left( {2; – 1; – 2} right) Rightarrow IJ = sqrt {{2^2} + {1^2} + {2^2}} = 3 > R = 1 Rightarrow J) nằm ở phía ngoài mặt cầu (left( S right)). Khi đó
(M{J_{max }} = IJ + R = 3 + 1 = 4)
Vậy ({T_{max }} = {3.4^2} + left( {{2^2} + {2^2} + {4^2}} right) + 2.left( {{1^2} + {1^2} + {2^2}} right) = 3.16 + 24 + 2.6 = 84).
Chọn C.