# A plane passes through a fixed point (p,q,r) and cut the axes in A, B, C. Show that the locus of the centre of the sphere OABC is p/x + q/y + r/z = 0 [closed]

I know how to find the equation of a sphere when four points, through which it passes is given. But here I'm unable to find coordinates of points $$A, B, C$$ in terms of $$p, q, r.$$

## closed as off-topic by Saad, YiFan, Thomas Shelby, Dbchatto67, LeucippusApr 19 at 5:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, YiFan, Thomas Shelby, Dbchatto67, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.

• Please check the question: it should be p/x + q/y + r/z = 2 – PTDS Apr 15 at 6:14
• You don't need to express $A,B,C$ in terms of $p,q,r$. Write the equation of the plane through $(p,q,r)$ whose normal vector is $(a,b,c)$, calculate $A,B,C$ in terms of $p,q,r,a,b,c$ and then find the center $(x,y,z)$ of the sphere. Then notice that $p/x+q/y+r/z=2$ (I agree with PTDS that it should be $2$) independently from $a,b,c$. – SMM Apr 15 at 8:02
• I confirm it's p/x + q/y + r/z = 0 – Manas Apr 16 at 7:54
• Can anyone give me complete solution – Manas Apr 16 at 7:55