Question

Let the equation of the sphere passing through the points $(0,0,0),(-1,1,1),(1,-1,1)$ and $(1,1,-1)$ be $k{x}^2+m{y}^2+n{z}^2=p(rx+y+z)$. Find $k+m+n+p-r$ ?

Answer 1

Equation of sphere passing through origin is given by,
$x^2+y^2+z^2+ax+by+cz=0$
Given this sphere passes through $(-1,1,1),(1,-1,1)$ and $(1,1,-1)$
$\Rightarrow 1+1+1-a+b+c=0\Rightarrow -a+b+c=-3$ ..... $\left(1\right)$
$1+1+1+a-b+c=0\Rightarrow a-b+c=-3$ ....... $\left(2\right)$
and $1+1+1+a+b-c=0\Rightarrow a+b-c=-3$ ....... $\left(3\right)$

Solving $\left(1\right),\left(2\right)$ and $\left(3\right),$ we get

$a=b=c=-3$
Thus, the required sphere is $x^2+y^2+z^2=3(x+y+z)$
$\Rightarrow k+m+n+p-r=5$