Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Given sphere is $\begin{array}{l}{x}^2+y^2+z^2-6x-12y-2z+20=0\\ \text{ Centre }\equiv (3,6,1)\end{array}$ Here, one end of diameter is $(2,7,3)$. Let the other end of the diameter be $(x,y,z)$ Centre of the sphere will be the mid-point of the ends of diameter. So, $(3,6,1)=\left(\frac{2+x}{2},\frac{7+y}{2},\frac{3+z}{2}\right)$ $\Rightarrow 2+\mathrm{x}=6\Rightarrow \mathrm{x}=4$ $\Rightarrow 7+\mathrm{y}=12\Rightarrow \mathrm{y}=5$ and $3+\mathrm{z}=2\Rightarrow \mathrm{z}=-$ and $3+z=2\Rightarrow z=-1,5,-1)$