Question: The intersection of the plane $\displaystyle x+2y+3z=10$ and the surface $\displaystyle z=x^2+y^2$ is an ellipse E. Find the points on E that are closest to the origin.

What I did:

First I join the two equations together to form: $\displaystyle x+2y+3x^2+3y^2=10$

Using $\displaystyle d^2=x^2+y^2+z^2$ and treating the above equation as a constraint:

$\displaystyle 2x=(1+6x)\lambda$

$\displaystyle 2y=(2+6y)\lambda$

$\displaystyle x+2y+3x^2+3y^2=10$

I managed to get: $\displaystyle (x=2/3,y=4/3),(x=-1,y=-2)$

Subst back into the equation I have $\displaystyle z= 20/9 \text{ and } 5$ respectively.

So the points I conclude is $\displaystyle (2/3,4/3,20/9)$

Is this right? Or do I have to form two contraint equations instead?