What I'm trying to do here is to establish the gradients of the ellipsoid and the plane. The plane has only one gradient (in any particular direction) whereas the ellipsoid gradient varies over the surface. Where the ellipsoid surface has the same gradient as the fixed gradient of the plane gives us point(s) on the surface. This means that the plane and the gradient (tangent) of the ellipsoid are parallel, when the gradient of the ellipsoid is measured in the same direction as the plane. The plane touches the ellipsoid only if they meet, that is, the coordinates of the ellipsoid at the surface match coordinates lying on the plane.
[In the following, I may not be able to give you the whole proof, due to my unfamiliarity with this type of problem, but I have an intuition regarding the essence of the solution. I hope that you will be able to glean enough for you to pursue the solution yourself.]
Let w=-sqrt(7)(1-2x^2-y^2)+2z^2. grad(w)=<4xsqrt(7),2ysqrt(7),4z>. This is derived from the differentials of the variable terms.
grad(w) at point P(a,b,c) is grad(w)|P=<4asqrt(7),2bsqrt(7),4c>, making the tangent plane:
4asqrt(7)(x-a)+2bsqrt(7)(y-b)+4c(z-c)=0 or 4asqrt(7)x+2bsqrt(7)y+4cz=4a^2sqrt(7)+2b^2sqrt(7)+4c^2.
The point P also lies on the plane: 3a+b-c=4, if the ellipsoid touches the plane. So, c=3a+b-4.
P must lie on the ellipsoid, so if a and b are set, c^2=z^2=(1-2a^2-b^2)sqrt(7)/2.
[However, if z=0, it is easier to find values for b and c on the surface of the ellipsoid. Such values are given by 2x^2+y^2=1, an ellipse itself. So P(a,sqrt(1-2a^2),0) is such a point. The tangential plane is 4asqrt(7)x+2sqrt(7(1-2a^2))y=4a^2sqrt(7)+2(1-2a^2) and 3a+sqrt(1-2a^2)-4=0 if P lies on the plane. So sqrt(1-2a^2)=4-3a must be satisfied. 1-2a^2=16-24a+9a^2, 11a^2-24a+15=0, which has no real solutions, so we cannot start with c=0, and we need to continue with the general case.]
With P(a,b,sqrt((1-2a^2-b^2)sqrt(7)/2)), (3a+b-4)^2=(1-2a^2-b^2)sqrt(7)/2, equating the squares of the z coords for the tangential plane and the given plane.
9a^2+6a(b-4)+b^2-8b+16=(1-2a^2-b^2)sqrt(7)/2.
9a^2+6ab-24+b^2-8b+16=sqrt(7)-a^2sqrt(7)-b^2sqrt(7)/2.
a^2(9+sqrt(7))+6ab+(b^2(1+sqrt(7)/2)-8b-8-sqrt(7))=0.
The discriminant is 36b^2-4(9+sqrt(7))(b^2(1+sqrt(7)/2)-8b-8-sqrt(7))=
36b^2-36b^2-36b^2sqrt(7)/2+288b+288+36sqrt(7)-4sqrt(7)b^2-14b^2+32sqrt(7)b+32sqrt(7)+28=
-4b^2(9sqrt(7)/2+sqrt(7)+7/2)+32b(9+sqrt(7))+316+68sqrt(7).
The discriminant must be positive for real a (for example if b=0).
[If b=0, P(a,0,sqrt((1-2a^2)sqrt(7)/2)) is the common point, (3a-4)^2=(1-2a^2)sqrt(7)/2.
9a^2-8a+16=(1-2a^2)sqrt(7)/2, a^2(9+sqrt(7))-8a+16-sqrt(7)/2=0.
The discriminant for this is 64-2(9+sqrt(7))(32-sqrt(7)) which is negative, so b cannot be 0.]
This "solution" is clearly becoming over-complicated and I suspect it's the wrong approach. Perhaps you will be able to take some of the ideas and develop them more elegantly. If I find a better method I will come back to this problem. Sorry for any inconvenience.