\[\boxed{\mathbf{676.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[O(0;0;0);\ \ A(4;0;0);\ \ B(0;6;0);\ \ \]
\[C(0;0; - 2).\]
\[\textbf{а)}\ \mathrm{\Delta}ABC - лежит\ в\ плоскости\ \]
\[Oxy;\]
\[D(x;y);\]
\[AD = BD = OD.\]
\[\left| \text{OD} \right| = \left| \overrightarrow{\text{OD}} \right| =\]
\[= \sqrt{(x - 0)^{2} + (y - 0)^{2}} =\]
\[= \sqrt{x^{2} + y^{2}};\]
\[\left| \text{AD} \right| = \left| \overrightarrow{\text{AD}} \right| =\]
\[= \sqrt{(x - 4)^{2} + (y - 0)^{2}} =\]
\[= \sqrt{x^{2} - 8x + 16 + y^{2}};\]
\[\left| \text{BD} \right| = \left| \overrightarrow{\text{BD}} \right| =\]
\[= \sqrt{(x - 0)^{2} + (y - 6)^{2}} =\]
\[= \sqrt{x^{2} + y^{2} - 12y + 36}.\]
\[x^{2} + y^{2} = x^{2} - 8x + y^{2} + 16 =\]
\[= x^{2} + y^{2} - 12y + 36\]
\[0 = - 8x + 16 = - 12y + 36\]
\[x = 2;\ \ y = 3.\]
\[R = \left| \text{OD} \right| = \sqrt{2^{2} + 3^{2}} = \sqrt{13}.\]
\[D(2;3;0) - центр\ окружности,\ \]
\[описанной\ около\ \mathrm{\Delta}ABC.\]
\[\textbf{б)}\ D(x;y;z) - равноудалена\ от\ \]
\[вершин\ тетраэдра:\]
\[OD = AD = BD = CD.\]
\[\left| \text{OD} \right| = \left| \overrightarrow{\text{OD}} \right| =\]
\[= \sqrt{(x - 0)^{2} + (y - 0)^{2} + (z - 0)^{2}} =\]
\[= \sqrt{x^{2} + y^{2} + z^{2}};\]
\[\left| \text{AD} \right| = \left| \overrightarrow{\text{AD}} \right| =\]
\[= \sqrt{(x - 4)^{2} + (y - 0)^{2} + (z - 0)^{2}} =\]
\[= \sqrt{x^{2} - 8x + 16 + y^{2} + z^{2}};\]
\[\left| \text{BD} \right| = \left| \overrightarrow{\text{BD}} \right| =\]
\[= \sqrt{(x - 0)^{2} + (y - 6)^{2} + (z - 0)^{2}} =\]
\[= \sqrt{x^{2} + y^{2} - 12y + 36 + z^{2}};\]
\[\left| \text{CD} \right| = \left| \overrightarrow{\text{CD}} \right| =\]
\[= \sqrt{(x - 0)^{2} + (y - 0)^{2} + (z + 2)^{2}} =\]
\[= \sqrt{x^{2} + y^{2} + z^{2} + 4z + 4}.\]
\[x^{2} + y^{2} + z^{2} =\]
\[= x^{2} - 8x + 16 + y^{2} + z^{2} =\]
\[= x^{2} + y^{2} - 12y + 36 + z^{2} =\]
\[= x^{2} + y^{2} + z^{2} + 4z + 4;\]
\[0 = - 8x + 16 = - 12y + 36 =\]
\[= 4z + 4\]
\[x = 2;\ \ y = 3;\ \ z = - 1.\]
\[D(2;3; - 1) - равноудалена\ от\ \]
\[вершин\ тетраэдра\ \text{OABC.}\]