Gọi (S) là mặt cầu ngoại tiếp tứ diện ABCD với A(0; 1; 0), B(4; 1; 0), C(2; 3; 0), D(2; 1; −2). Khẳng định nào sau đây là đúng?

A. x² + y² + z² – 2x + 4y – 4z = 0;

B. (x − 1)² + (y − 2)² + (z − 2)² = 9;

C. (x − 2)² + (y − 4)² + (z − 4)² = 20;

D. x² + y² + z² + 2x – 4y + 4z = 9.

A. (x – 1)² + (y − 1)² + (z – 2)² = 4;

B. (x + 1)² + (y + 1)² + (z + 2)² = 4;

C. (x − 1)² + (y − 1)² + (z − 2)² = 2;

D. (x + 1)² + (y + 1)² + (z + 2)² = 4.

A. x² + y² + z² + 5x + 5y − 5z – 6 = 0;

B. \({x^2} + {y^2} + {z^2} + \frac{5}{2}x + \frac{5}{2}y - \frac{5}{2}z - 6 = 0\);

C. x² + y² + z² +5x + 5y − 5z + 6 = 0;

D. \({x^2} + {y^2} + {z^2} + \frac{5}{2}x + \frac{5}{2}y - \frac{5}{2}z + 6 = 0\).

A. x² + y² + z² + 3x + y − z – 6 = 0;

B. (S) có tâm \(I\left( {\frac{{ - 3}}{2}; - \frac{1}{2};\frac{1}{2}} \right)\);

C. (S) có tâm \(I\left( {\frac{3}{2};\frac{1}{2}; - \frac{1}{2}} \right)\);

D. (S) có bán kính \(R = \frac{{\sqrt {35} }}{2}\).

A. x² + y² + z² + x + y + z – 2 = 0;

B. x² + y² + z² − x − y – z + 2 = 0;

C. x² + y² + z² − x − y – z − 2 = 0;

D. x² + y² + z² + x + y + z − 6 = 0.

A. \({\left( {x + \frac{1}{2}} \right)^2} + {\left( {y + \frac{1}{2}} \right)^2} + {\left( {z + \frac{1}{2}} \right)^2} = \frac{3}{4}\);

B. \({\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} + {\left( {z - \frac{1}{2}} \right)^2} = \frac{{\sqrt 3 }}{2}\);

C. \({\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} + {\left( {z - \frac{1}{2}} \right)^2} = \frac{3}{4}\);

D. \({\left( {x + \frac{1}{2}} \right)^2} + {\left( {y + \frac{1}{2}} \right)^2} + {\left( {z + \frac{1}{2}} \right)^2} = \frac{{\sqrt 3 }}{2}\).

A. \({x^2} + {y^2} + {z^2} + \frac{1}{7}x + \frac{{15}}{7}y - \frac{{37}}{7}z = 0\);

B. \({x^2} + {y^2} + {z^2} + \frac{1}{{14}}x + \frac{{15}}{{14}}y - \frac{{37}}{{14}}z = 0\);

C. \({x^2} + {y^2} + {z^2} + \frac{1}{7}x - \frac{{15}}{7}y + \frac{{37}}{7}z = 0\);

D. \({x^2} + {y^2} + {z^2} - \frac{1}{{14}}x - \frac{{15}}{{14}}y + \frac{{37}}{{14}}z = 0\).