Trong không gian Oxyz,cho hai điểm \[A\left( {1; - 3;2} \right),{\rm{ }}B\left( {2; - 2;3} \right).\] Tìm tọa độ điểm K đối xứng với A qua B.

A.\[P = 2020.\]

B.\[P = 2020!.\]

C.\[P = \frac{1}{{2020}}.\]

D.\[P = 1.\]

A.\[\Delta:\frac{{x - 2}}{2} = \frac{y}{1} = \frac{{z - 2}}{{ - 1}}.\]

B.\[\Delta:\frac{{x - 2}}{2} = \frac{y}{{ - 5}} = \frac{{z - 2}}{{ - 1}}.\]

C.\[\Delta:\frac{{x - 3}}{3} = \frac{{y - 1}}{1} = \frac{{z - 1}}{1}.\]

D.\[\Delta:\frac{{x - 3}}{2} = \frac{{y - 1}}{{ - 5}} = \frac{{z - 1}}{{ - 1}}.\]

A.\[\vec u = \left( {2;3;1} \right).\]

B.\[\vec u = \left( {2;1; - 2} \right).\]

C.\[\vec u = \left( {2; - 3;1} \right).\]

D.\[\vec u = \left( {2;1;2} \right).\]

A.\[y = {x^3} - 3{x^2} + 3x + 1.\]

B.\[y = - {x^3} + 3{x^2} + 1.\]

C.\[y = {x^3} - 3x + 4.\]

D.\[y = - {x^3} - 3{x^2} - 1.\]

A.\[\left\{ {\begin{array}{*{20}{l}}{x = 1 - 3t}\\{y = 0}\\{z = 1 + t}\end{array}} \right..\]

B.\[\left\{ {\begin{array}{*{20}{l}}{x = 1 - 3t}\\{y = 0}\\{z = 1 - t}\end{array}} \right..\]

C.\[\left\{ {\begin{array}{*{20}{l}}{x = 1 - 3t}\\{y = t}\\{z = 1 + t}\end{array}} \right..\]

D.\[\left\{ {\begin{array}{*{20}{l}}{x = 1 + 3t}\\{y = 0}\\{z = 1 + t}\end{array}} \right..\]

A.\[V = \frac{{128\pi }}{3}.\]

B.\[V = 128\pi .\]

C.\[V = \frac{{256\pi }}{3}.\]

D.\[V = 96\pi .\]