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.
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} &amp;= 1 \\
  \frac{25}{2}k^{2} &amp;= 1 \\
  k &amp;= \pm \frac{\sqrt{2}}{5} \\
  (X,Y,Z) &amp;=
  \left(
    \pm \frac{3\sqrt{2}}{5},
    \mp \frac{1}{5\sqrt{2}},
    \pm \frac{\sqrt{2}}{5} \right)
\end{align*}
Consider the function $f(x,y,z) = x^2+2y^2+3z^2-1$ then the $E$=ellipsoid is given by $f^{-1}(0)$ i.e a level surface. Let $p \in E$ then $\nabla f(p)$ is a normal vector for the tangent plane. For the tangent plane to be parallel to $3x-y+3z=1$, you need $p \in E$ such that $\nabla f(p) = \lambda \langle 3,-1,3\rangle$.