Plane $3x + y - z= 4$ touches the ellipsoid $2z^2 = \sqrt7(1 - 2x^2 -y^2)$ Show that the Plane $3x + y - z= 4$ touches the ellipsoid $2z^2 = \sqrt7(1 - 2x^2 -y^2)$

My attempt: First I tried to convert the equation of ellipsoid in general form and then further applying the condition of plane as tangent which is $$ \frac{u^2}{a} + \frac{v^2}{b} + \frac{w^2}{c} = p^2$$
Rewriting the equation of conic we get $$2x^2 + y^2 + \frac{2}{\sqrt7}z^2 = 1$$
Using this we get $a = 2 , b = 3$ and $c = \frac{2}{\sqrt7}$
and we using equation of plane we get $u = 3 , v = 1 , w = -1$ and $p = 4$
The condition is not satisfied. What am I doing wrong here? I seem to do everything correct and tried this question five times.
I'm really stuck at this problem. Kindly help.
Note: Please ignore the B3P36 in the image that was written by me as a code to something. It has no significance to the problem.
 A: The equation of the plane in general form is
$$3 x + y - z - 4 = 0$$
and therefore its normal is $(3,1,-1)$. Its minimum distance $D$ to origin is
$$D = \frac{\lvert -4 \rvert}{\sqrt{3^2 + 1^2 + (-1)^2 }} = \frac{4}{\sqrt{11}} \approx 1.206$$
and the unit normal vector is $\hat n = (3/\sqrt{11},1/\sqrt{11},-1/\sqrt{11})$. Distance $t$ from origin along the plane normal is therefore
$$\vec n (t) = \left ( \frac{3 t}{\sqrt{11}}, \frac{t}{\sqrt{11}}, -\frac{t}{\sqrt{11}} \right )\tag{1}\label{1}$$
Divide both sides of the formula for the ellipsoid
$$2 z^2 = \sqrt{7} ( 1 - 2 x^2 - y^2 )$$
by $\sqrt{7}$ and move $-2 x^2 - y^2$ to the other side: 
$$\frac{2}{\sqrt{7}} z^2 + 2 x^2 + y^2 = 1\tag{2}\label{2}$$
If we move the coefficients to denominators, we can see the general form:
$$\frac{x^2}{(\sqrt{1/2})^2} + \frac{y^2}{1^2} + \frac{z^2}{(\sqrt{\sqrt{7}/2})^2} = 1$$
and we see it is an axis-aligned ellipsoid with semi-principal axes $a = \sqrt{1/2} \approx 0.7071$, $b = 1$, and $c = \sqrt{\sqrt{7}/2} \approx 1.150$.
(At this point, we notice that all the semi-principal axes are smaller than the minimum distance from the plane to the center of the ellipsoid, origin, which means the two surfaces cannot come into contact. The ellipsoid is just too small.) 
The closest the ellipsoid comes to the plane (if they do not touch), the point that contacts the plane (if they contact at exactly one point), or the point on the ellipsoid furthest on the other side of the plane (if they intersect in a ring), is on the line parallel to the plane normal that passes through the center of the ellipsoid (here, origin).
So, to find that point on the ellipsoid we insert equation $\eqref{1}$ into $\eqref{2}$:
$$\frac{2}{\sqrt{7}} \left ( -\frac{t}{\sqrt{11}} \right )^2 + 2 \left ( \frac{3 t}{\sqrt{11}} \right )^2 + \left ( \frac{t}{\sqrt{11}} \right )^2 = 1$$
$$\frac{2}{11 \sqrt{7}} t^2 + \frac{18}{11} t^2 + \frac{1}{11} t^2 - 1 = 0$$
and solve for $t$, the (signed) distance from origin (negative referring to opposite direction from plane surface normal):
$$t = \pm \frac{\sqrt{10241 + 154\sqrt{7}}}{2\sqrt{7} + 133} \approx \pm 0.7462$$
Because the closest point to origin the plane comes to is $\vec n(D)$, i.e. at distance $D \approx 1.206$, the plane never comes closer than approximately $1.206 - 0.7462 = 0.4598$ to the ellipsoid.
