Trong không gian Oxyz, cho đường thẳng (d) đi qua  \[{M_0}\left( {{x_0},{y_0},{z_0}} \right)\;\;\]và nhận \[\overrightarrow u = \left( {a,b,c} \right),\;\;{a^2} + {b^2} + {c^2} > 0\;\]làm một vecto chỉ phương. Hãy chọn khẳng định sai trong bốn khẳng định sau?

A.(0;1;2)           

B.(1;0;1)

C.(2;−2;1)        

D.(3;−4;1) 

A.\[d:\frac{{x - 1}}{4} = \frac{{y - 2}}{{ - 7}} = \frac{{z - 3}}{{ - 1}}\]

B. \[d:\frac{{x - 1}}{4} = \frac{{y - 2}}{7} = \frac{{z - 3}}{1}\]

C. \[d:\frac{{x - 1}}{{ - 4}} = \frac{{y - 2}}{{ - 7}} = \frac{{z - 3}}{1}\]

D. \[d:\frac{{x - 1}}{4} = \frac{{y - 2}}{{ - 7}} = \frac{{z - 3}}{1}\]

A.(−1;−1;1)     

B.(−1;1;1)        

C.(0;1;1)

D.(0;1;0)

A.\[\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}\]

B. \[\frac{{x - {x_0}}}{{ - a}} = \frac{{y - {y_0}}}{{ - b}} = \frac{{z - {z_0}}}{{ - c}}\]

C. \[\frac{{x + {x_0}}}{a} = \frac{{y + {y_0}}}{b} = \frac{{z + {z_0}}}{c}\]

D. \[\frac{{x + {x_0}}}{a} = \frac{{y + {y_0}}}{{ - b}} = \frac{{z + {z_0}}}{c}\]

A.\(\left\{ {\begin{array}{*{20}{c}}{x = 6t}\\{y = - 4t}\\{z = - 3t}\end{array}} \right.\)

B. \(\left\{ {\begin{array}{*{20}{c}}{x = 6t}\\{y = 2 + 4t}\\{z = - 3t}\end{array}} \right.\)

C. \(\left\{ {\begin{array}{*{20}{c}}{x = 6t}\\{y = 4t}\\{z = - 3t}\end{array}} \right.\)

D. \(\left\{ {\begin{array}{*{20}{c}}{x = 6t}\\{y = 4t}\\{z = 1 - 3t}\end{array}} \right.\)

A.\[\left( {a;b;c} \right)\]

B. \[\left( {a;b;c} \right)\]

C. \[\left( {{x_0};{y_0};{z_0}} \right)\]

D. \[\left( { - {x_0}; - {y_0}; - {z_0}} \right)\]