공간좌표

Call the centre of the circumscribing circle $(\bar x,\bar y)$. This centre must be equidistant from the three vertices. So

$$\bar x^2+\bar y^2=(\bar x-a)^2+\bar y^2=(\bar x-b)^2+(\bar y-c)^2$$

or, subtracting $\bar x^2+\bar y^2$ from the three equal expressions,

$$0=a^2-2a\bar x=b^2-2b\bar x+c^2-2c\bar y$$

which implies

$$\bar x=\frac{a}{2}\qquad\qquad \bar y =\frac{b^2+c^2-2b\bar x}{2c}=\frac{b^2+c^2-ab}{2c}$$

The radius is the distance from the vertex $(0,0)$ to the centre $(\bar x,\bar y)$, which is $\sqrt{\big(\frac{a}{2}\big)^2+\big(\frac{b^2+c^2-ab}{2c}\big)^2}$.

The distance from $P$ to the point $(0,0,1)$ is $\sqrt{x^2+y^2+(z-1)^2}$. The distance from $P$ to the specified plane is $|z+1|$. Hence the equation of the surface is

$$x^2+y^2+(z-1)^2=(z+1)^2\text{ or } x^2+y^2=4z$$

All points on this surface have $z\ge 0$. The set of points on the surface that have any fixed value, $z_0\ge 0$, of $z$ consists of a circle that is centred on the $z$–axis, is parallel to the $xy$-plane and has radius $2\sqrt{z_0}$. The surface consists of a stack of these circles, starting with a point at the origin and with radius increasing vertically. The surface is a paraboloid and is sketched below.

Let $(x,y,z)$ be a point in $P$. The distances from $(x,y,z)$ to $(3,-2,3)$ and to $(3/2,1,0)$ are

$$\sqrt{(x-3)^2+(y+2)^2+(z-3)^2}\quad\text{ and }\quad \sqrt{(x-3/2)^2+(y-1)^2+z^2}$$

respectively. To be in $P$, $(x,y,z)$ must obey

$$\begin{aligned} \sqrt{(x-3)^2+(y+2)^2+(z-3)^2}&=2\sqrt{(x-3/2)^2+(y-1)^2+z^2} \\ (x-3)^2+(y+2)^2+(z-3)^2&=4(x-3/2)^2+4(y-1)^2+4z^2 \\ x^2-6x+9+y^2+4y+4+z^2-6z+9&=4x^2-12x+9+4y^2-8y+4+4z^2 \\ 3x^2-6 x+3y^2-12y+3z^2+6z-9&=0 \\ x^2-2 x+y^2-4y+z^2+2z-3&=0 \\ (x-1)^2+(y-2)^2+(z+1)^2&=9 \end{aligned}$$

This is a sphere of radius 3 centered on $(1,2,-1)$.