<s>[INST] Find the points on the ellipsoid $x^2 + 2y^2 + 3z^2 = 1$ where the tangent plane is parallel to the plane $3x - y + 3z = 1$. Find the points on the ellipsoid $x^2 + 2y^2 + 3z^2 = 1$, where the tangent plane is parallel to the plane $3x - y + 3z = 1$.
I'm not sure how to go about solving this. I'd appreciate some help.
 [/INST] Let the contact point be $(X,Y,Z)$, now the tangent plane is

$$Xx+2Yy+3Zz=1$$

Comparing coefficients, 
$$X:2Y:3Z=3:-1:3$$
That is $$\frac{X}{3}=\frac{2Y}{-1}=\frac{3Z}{3}=k$$
Now substitute $(X,Y,Z)=\left( 3k,-\dfrac{k}{2},k \right)$ into $x^{2}+2y^{2}+3z^{2}=1$
We have
\begin{align*}
  (3k)^{2}+2\left( -\frac{k}{2} \right)^{2}+3(k)^{2} &= 1 \\
  \frac{25}{2}k^{2} &= 1 \\
  k &= \pm \frac{\sqrt{2}}{5} \\
  (X,Y,Z) &=
  \left(
    \pm \frac{3\sqrt{2}}{5},
    \mp \frac{1}{5\sqrt{2}},
    \pm \frac{\sqrt{2}}{5} \right)
\end{align*}
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