Đáp án+Giải thích các bước giải:

$14)$

$I$ là tâm mặt cầu

$I \in (Oxy) \Rightarrow I(a;b;0)$

$A,B,C$ thuộc mặt cầu

$\Rightarrow \left\{\begin{array}{l} IA^2=IB^2\\IA^2=IC^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} (a-2)^2+(b-5)^2+3^2=(a-10)^2+(b-5)^2+3^2\\(a-2)^2+(b-5)^2+3^2=(a-6)^2+(b-11)^2+8^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} a=6\\b=\dfrac{45}{4}\end{array} \right.\\ R^2=(6-2)^2+\left(\dfrac{45}{4}-5\right)^2+3^2=\dfrac{1025}{16}$

Phương trình mặt cầu: 

$(x-6)^2+\left(y-\dfrac{45}{4}\right)^2+z^2=\dfrac{1025}{16}$

$15)$

$I(a;b;c)$ là tâm mặt cầu

$A,B,C$ thuộc mặt cầu

$\Rightarrow \left\{\begin{array}{l} IA^2=IB^2\\IA^2=IC^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} (a+2)^2+(b-4)^2+(c-1)^2=(a-3)^2+(b-1)^2+(c+3)^2\\(a+2)^2+(b-4)^2+(c-1)^2=(a+5)^2+b^2+c^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 5 a - 3 b - 4 c + 1=0\\3 a + 4 b + c + 2=0\end{array} \right.(1)\\ I \in (\alpha): 2x+y-z+3=0 \Rightarrow 2a+b-c+3=0(2)\\ (1)(2) \Rightarrow I=(1; -2; 3)\\ R^2=(1+5)^2+(-2)^2+3^2=49$

Phương trình mặt cầu: 

$(x-1)^2+(y+2)^2+(z-3)^2=49\\ 16)$

$I(a;b;c)$ là tâm mặt cầu

$A,B,C,D$ thuộc mặt cầu

$\Rightarrow \left\{\begin{array}{l} IA^2=IB^2\\IA^2=IC^2\\ IA^2=ID^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} (a-1)^2+b^2+(c-2)^2=(a-2)^2+(b+1)^2+(c-1)^2\\(a-1)^2+b^2+(c-2)^2=a^2+(b-2)^2+(c-1)^2\\ (a-1)^2+b^2+(c-2)^2=(a+1)^2+(b-3)^2+c^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} a = -\dfrac{7}{2}\\ b = -\dfrac{5}{2}\\ c = -\dfrac{3}{2}\end{array} \right.\\ R^2=\left(-\dfrac{7}{2}-1\right)^2+\left(-\dfrac{5}{2}\right)^2+\left(-\dfrac{3}{2}-2\right)^2=\dfrac{155}{4}$

Phương trình mặt cầu: 

$\left(x+\dfrac{7}{2}\right)^2+\left(y+\dfrac{5}{2}\right)^2+\left(z+\dfrac{3}{2}\right)^2=\dfrac{155}{4}.$