Let `(alpha, beta, gamma)` be any point on the locus. Then according to the given condition, `(alpha, beta, gamma)` is the centre of the sphere through the origin. Therefore, its equation is given by
`" "(x-alpha)^(2)+(y-beta)^(2)+(z-gamma)^(2)=(0-alpha)^(2)+(0-beta)^(2)+(0-gamma)^(2)`
`" "x^(2)+y^(2)+z^(2)-2alphax-2betay-2gammaz=0`
To obtain its point of intersection with the x-axis, we put y=0 and z=0, so that
`" "x^(2)-2alphax=0 or x(x-2alpha)=0 or x=0, 2alpha`
Thus, the plane meets the x-axis at O (0, 0, 0) and `A(2alpha, 0, 0)`. Similarly, it meets the y-axis at `O(0, 0, 0) and B(0, 2beta, 0)`, and the z-axis at `O(0, 0, 0) and C(0, 0, 2gamma)`.
The equation of the plane through `A, B and C` is
`" "(x)/(2alpha)+(y)/(2beta)+(z)/(2gamma) =1" "`(intercept form)
Since it passes through `(a, b, c)`, we get
`" "(a)/(2alpha)+(b)/(2beta)+(c)/(2gamma)=1 or (a)/(alpha)+(b)/(beta)+(c)/(gamma)=2`
Hence, locus of `(alpha, beta, gamma)` is `(a)/(x)+(b)/(y)+(c)/(z)=2`