what is the shortest distance between the spheres s (𝑥 − 1)2 + (𝑦 + 1)2 + 𝑧2 = 6 and (𝑥 + 1)2 + 𝑦2 + (𝑧 + 1)2 = 15?

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Given spheres (x-1)^{2}+(y+1)^{2}+z^{2}=6(x+1)^{2}+y^{2}+(z+1)^{2}=15centre of the first sphere C_{1}=(1,-1,0)Radius of the first sphere r_{1}=\sqrt{6}centre of the second sphere C_{2}=(-1,0,-1)Radius of the second sphere r_{2}=\sqrt{15}The distance between to centres =\sqrt{(1+1)^{2}+(-1-0)^{2}+(0+1)^{2}}=\sqrt{4+1+1}=\sqrt{6}The distance between the two spheres is equal to\begin{array}{l}=(\text { The distance between centres of two spheres }) \text {-(sum of the radius of two spheres }) \\=\sqrt{6}-(\sqrt{6}+\sqrt{15}) \\=-\sqrt{15}<0 \\\text { i.e } C_{1} C_{2}<\left(\mathrm{r}_{1}+r_{2}\right)\end{array}Hence the above two spheres are intersect\therefore The shortest distance between two spheres is 0 . ...