Lagrange multipliers from hell I was asked to solve this question, decided to try and solve it with lagrange multipliers as I see no other way:
"Find the closest and furthest points on the circle made from the intersection of the ball $(x-1)^2+(y-2)^2+(z-3)^2=9$ and the plane $x-2z=0$ from the point $(0,0)$".
What I did:
the distance for any point $(x,y,z)$ from the origin is $d(x,y,z)=\sqrt{x^2+y^2+z^2}$. so using lagrange multipliers we have:
$d(x,y,z)=\sqrt{x^2+y^2+z^2}$
$C_1(x,y,z)=(x-1)^2+(y-2)^2+(z-3)^2-9$
$C_2(x,y,z)=x-2z$
$L(x,y,z) = d-\lambda_1C_1-\lambda_2C_2 $ meaning:
$L(x,y,z)=\sqrt{x^2+y^2+z^2}-\lambda_1[(x-1)^2+(y-2)^2+(z-3)^2-9]-\lambda_2(x-2z)$
Let's derive and solve when derivatives are zero:
$\frac{\partial L}{\partial x}= \frac{x}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(x-1)-\lambda_2=0$
$\frac{\partial L}{\partial y} = \frac{y}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(y-2)=0$
$\frac{\partial L}{\partial z} = \frac{z}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(z-3)+2\lambda_2=0$
$\frac{\partial L}{\partial \lambda_1} = -(x-1)^2-(y-2)^2-(z-3)^2+9=0$
$\frac{\partial L}{\partial \lambda_2} = 2z-x=0$
Solving this monstrous system seems very unlikely, and very difficult, and not how the question is meant to be solved. am I missing something?

$\newcommand{\e}{\mathbf{e}}$Here's an algebraic approach:
Let $P$ denote the plane with equation $x - 2z = 0$ and $S$ the sphere
$$
(x - 1)^{2} + (y - 2)^{2} + (z - 3)^{3} = 9.
$$
The vectors $\e_{1} = (2, 0, 1)/\sqrt{5}$ and $\e_{2} = (0, 1, 0)$ are an orthonormal basis of $P$. The orthogonal projection to $P$ of $c = (1, 2, 3)$, the center of $S$, is
$$
\langle c, \e_{1}\rangle \e_{1} + \langle c, \e_{2}\rangle \e_{2}
  = \sqrt{5}\e_{1} + 2\e_{2}
  = (2, 2, 1)
  = c_{0}.
$$
Since the distance from $c$ to $c_{0}$ is $\sqrt{5}$, the intersection of $S$ and $P$ is a circle of radius $\sqrt{3^{2} - 5} = 2$ centered at $c_{0}$.
In the Cartesian coordinates defined by $\{\e_{1}, \e_{2}\}$, the center $c_{0}$ has coordinates $(\sqrt{5}, 2)$, and a bit of algebra shows the nearest point to the origin is $(\sqrt{5}, 2)/3$, while the furthest point is $5(\sqrt{5}, 2)/3$. 
Converting back to spatial coordinates, the nearest point to the origin is $(2/3, 2/3, 1/3)$ and the furthest point is $(10/3, 10/3, 5/3)$.
