Finding a point on an ellipsoid Find a point on the ellipsoid $x^2+4y^2+z^2=9$ where the tangent plane is perpendicular to the line with parametric equations \begin{align}x&=2+2t\\y&=1+2t\\z&=3-t\end{align}
The answer to this question is: $$\left(\frac{6\sqrt{13}}{13}, \frac{6\sqrt{13}}{13}, -\frac{3\sqrt{13}}{13}\right)\\
\left(-\frac{6\sqrt{13}}{13}, -\frac{6\sqrt{13}}{13}, \frac{3\sqrt{13}}{13}\right)$$
I know the normal is $(x, 4y, z)$. What step comes after this?
Thanks in advance. 

The normal to the plane must be parallel to the given line, hence
$$(x,4y,z)=\lambda(2,2,-1).$$
Combining with the equation of the ellipsoid,
$$x^2+4y^2+z^2=9,$$
we get
$$2^2\lambda^2+4\frac1{2^2}\lambda^2+(-1)^2\lambda^2=6\lambda^2=9,$$ and solve for $\lambda$.
$$(x,y,z)=\pm\sqrt{\frac32}\left(2,\frac12,-1\right).$$
