The equation of the sphere through the points $(1,0,0),(0,1,0)$ and $(1,1,1)$ and having the smallest radius
The equation of the sphere through the points $(1,0,0),(0,1,0)$ and $(1,1,1)$ and having the smallest radius
$3\left(x^2+y^2+z^2\right)-4 x-4 y-2 z+1=0$
$2\left(x^2+y^2+z^2\right)-3 x-3 y-z+1=0$
$x^2+y^2+z^2-x-y+z+1=0$
$x^2+y^2+z^2-2 x-2 y+4 z+1=0$
Solution
Given points are $A(1,0,0), B(0,1,0)$ and $C(1,1,1)$
$\begin{aligned} & A B=\sqrt{(0-1)^2+(1-0)^2+0^2}=\sqrt{2} \\ & B C=\sqrt{(0-1)^2+0^2+1^2-\sqrt{2}} \\ & C A=\sqrt{0^2+1^2+1^2}=\sqrt{2} \\ & \end{aligned}$
$\therefore M A C$ is an equilateral triangle.
$\therefore$ Centre of sphere $=$ Centroid of $\triangle A B C$
$=C^{\prime}\left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)$
$\therefore$ Radius of sphere $A C^{\prime}$
$\begin{aligned} & =\sqrt{\left(\frac{2}{3}-1\right)^2+\left(\frac{2}{3}\right)^2+\left(\frac{1}{3}\right)^2} \\ & =\sqrt{\frac{1}{9}+\frac{4}{9}+\frac{1}{9}}=\frac{1}{3} \sqrt{6}\end{aligned}$
$\therefore$ Equation of sphere is
$\begin{gathered}\left(x-\frac{2}{3}\right)^2+\left(y-\frac{2}{3}\right)^2+\left(z-\frac{1}{3}\right)^2=\left(\frac{\sqrt{6}}{3}\right)^2 \\ \Rightarrow x^2+y^2+z^2-\frac{4 x}{3}-\frac{4 y}{3}-\frac{2 z}{3}+\frac{4}{9}+\frac{4}{9}+\frac{1}{9} \\ =\frac{6}{9}=\frac{2}{3} \\ \Rightarrow 3\left(x^2+y^2+z^2\right)-4 x-4 y-2 z+1=0\end{gathered}$