Being given a general quadric with implicit equation:
$$\tag{1}ax^2+by^2+cz^2+2dxy+2eyz+2fzx+2gx+2hy+2iz+k=0,$$
the key point is that the tangent plane to this quadric in $(x',y',z')$ is obtained by "polarization", i.e., by looking at the homogeneous version of (1). This homogeneous form is obtained by introducing a 4th variable $t$, atop variables $x,y,z$, in such a way that if a term is only 1st degree, it is multiplied by $t$, and if it is a constant, it is multiplied by $t^2$: thus, all terms become of degree 2 (whence the name "homogenization"), in this way:
$$\tag{1}ax^2+by^2+cz^2+2dxy+2eyz+2fzx+2gtx+2hty+2itz+kt^2=0$$
Remark: if $t=1$, one finds back (1).
Then the method amounts to build the bilinear form associated with this quadratic form in 4 variables:
$\tag{2}axx'+byy'+czz'+d(x'y+xy')+e(y'z+yz')+f(z'x+zx')+g(t'x+x't)+h(t'y+ty')+i(t'z+tz')+ktt'=0.$
Remark: if all the "primes" are suppressed, we are back to equation (2).
Then make $t=t'=1$ in (2),
$$\tag{3}axx'+byy'+czz'+d(x'y+xy')+e(y'z+yz')+f(z'x+zx')+g(x+x')+h(y+y')+i(z+z')+k=0,$$
We have obtained in (3) the equation of the tangent plane to the quadric at point $(x',y',t')$.
Remark: this techniques can be justified in the framework of projective geometry.
Applying this to our case, we get, for the tangent plane at $(x_0,y_0,z_0)$ is
$$xx_0+4yy_0+zz_0=9$$
This is the equation of a plane with normal vector
$$\pmatrix{x_0\\4y_0\\z_0} \text{which is desired to be proportional to} \pmatrix{-4\\8\\-2}.$$
Thus, we must look for a coefficient $\alpha$ such that:
$$\tag{4}\pmatrix{x_0\\4y_0\\z_0}=\pmatrix{-4\alpha\\8\alpha\\-2\alpha}$$
Point $(x_0,y_0,z_0)$ is a point of the ellipsoid if and only if:
$$(-4\alpha)^2+4(2\alpha)^2+(-2\alpha)^2=9$$
i.e., $$\alpha^2=\frac{9}{36}=\frac{1}{4} \ \ \Leftrightarrow \ \ \alpha=\pm \frac{1}{2}$$
Whence the two solution points by plugging these opposite values of $\alpha$ in (4):
$$\pmatrix{x_0\\y_0\\z_0}=\pm \frac{1}{2} \pmatrix{-4\\2\\-2}=\pm \pmatrix{-2\\1\\-1}.$$
Best Answer
Since the two planes are perpendicular, the normal vector of the second plane (i.e. $(1,2,-1)$) is parallel to the first. To find the normal vector of the first plane, take the cross product: $(1,2,-1)×(-3,-3,0)=(-3,3,3)$. Finding the constant in the plane equation is now a matter of substitution: $k=-3\cdot6+3\cdot2+3\cdot1=-9$. So the plane's equation is $-3x+3y+3z=-9$ or $x-y-z=3$.