Intercept Form of a Plane

Let the intercepts of the variable plane on the $$x, y,$$ and $$z$$ axes be $$\alpha, \beta,$$ and $$\gamma$$ respectively.
The equation of the plane is:
$$\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1$$
Since it passes through the fixed point $$(a, b, c)$$, we have:
$$\frac{a}{\alpha} + \frac{b}{\beta} + \frac{c}{\gamma} = 1 $${--- (Equation 1)}
The coordinates of the intercepts are $$A(\alpha, 0, 0)$$, $$B(0, \beta, 0)$$, and $$C(0, 0, \gamma)$$.
The equation of a sphere passing through the origin $$O(0,0,0)$$ and points $$A, B, C$$ is:
$$x^2 + y^2 + z^2 - \alpha x - \beta y - \gamma z = 0$$
The center $$(x_1, y_1, z_1)$$ of this sphere is given by:
$$x_1 = \frac{\alpha}{2}, \quad y_1 = \frac{\beta}{2}, \quad z_1 = \frac{\gamma}{2}$$
Therefore, $$\alpha = 2x_1$$, $$\beta = 2y_1$$, and $$\gamma = 2z_1$$.
Substitute these values of $$\alpha, \beta, \gamma$$ into Equation 1:
$$\frac{a}{2x_1} + \frac{b}{2y_1} + \frac{c}{2z_1} = 1 \implies \frac{a}{x_1} + \frac{b}{y_1} + \frac{c}{z_1} = 2$$
Replacing $$(x_1, y_1, z_1)$$ with general coordinates $$(x, y, z)$$ gives the locus.