Question

Центр, R=?

$x {}^{2} + y {}^{2} + z {}^{2} - x + 2y - 3z - 4 = 6$

Answer 1

$( {x - \frac{1}{2} })^{2} + ( {y + 1})^{2} + ( {z - \frac{3}{2} })^{2} = \frac{1}{4} + 1 + \frac{9}{4} + 4 + 6 \\ ( {x - \frac{1}{2} })^{2} + ( {y + 1})^{2} + ( {z - \frac{3}{2} })^{2} = \frac{27}{2} \\ ( {x - \frac{1}{2} })^{2} + ( {y + 1})^{2} + ( {z - \frac{3}{2} })^{2} = ({ \frac{3 \sqrt{3} }{ \sqrt{2} } })^{2}$
Ответ: { \frac{3 \sqrt{3} }{ \sqrt{2} } }