I have seen proofs that utilize a geometric approach by constructing perpendicular segments, etc. but yet no purely algebraic approach.
Let's consider for simplicity that the sphere is $x^2+y^2+z^2=R^2$ and the plane is $ax+by+cz+d=0$ with the appropriate constraints, such that the plane intersects the sphere and $(a,b,c)\neq (0,0,0)$.
Let $t=-\frac d{a^2+b^2+c^2}$. Then $(x_0,y_0,z_0)=(at,bt,ct)$ is a point in the plane. If $(x,y,z)$ is in thge plane and also on the sphere, then $$\begin{align}(x-x_0)^2+(y-y_0)^2+(z-z_0)^2&=x^2+y^2+z^2+x_0^2+y_0^ 2+z_0^2\\&\quad -2(x_0x+y_0y+z_0z)\\&=R^2+x_0^2+y_0^2+z_0^2-2t(ax+by+cz)\\&=R^2++x_0^2+y_0^2+z_0^2+2td\\&=\text{const.} \end{align}$$