Let's split the solution into several steps. Suppose your surface $S$ is given by the equation $F(x)=0$ with $F$ being, let's say, $C^1$ function.
The tangent space $T_xS$ to the surface $S$ at the point $x$ is orthogonal to $\nabla F(x)$. One of equivalent definition of $T_xS$ is the following:
Let $\gamma:[-1,1]\to S$ be a $C^1$ map (in the sense of $\Bbb R\to \Bbb R^3$, for example) such that $\gamma(0)=x$. Then we consider the vector $\frac {d}{dt}\gamma(0)$ and take all possible $\gamma$ - and, therefore, all possible $\frac {d}{dt}\gamma(0)$ form $T_xS$. Now consider any such $\gamma$ and study $F(\gamma(t)) = 0$ (because $\forall t$ we have $\gamma(t)\in S$). By deriving this expressions with respect to $t$ at $t=0$ we obtain $\nabla F(x)\cdot \frac {d}{dt}\gamma(0) =0$, which is, in other words, $\nabla F(x)\bot T_xS$.
In your case, the function $F$ is $ax^2 + by^2 + cz^2 - 1$. The tangent space is given by the equation $\alpha x+\beta y+\theta z+\phi=0$. This plane should be orthogonal to the vector $\nabla (ax^2 + by^2 + cz^2 -1)\big|_{(x,y,z)=( x_0, y_0, z_0)} = 2(ax_0,by_0,cz_0)$; but already this plane is orthogonal to the vector $(\alpha,\beta,\theta)$. As the plane can be orthogonal to only one vector (up to a constant factor), we obtain that the plane is given by $2a x_0x+2by_0y+2cz_0z+\phi=0$.
In order to find $\phi$, we recall that the point $( x_0, y_0, z_0)$ lies both on the ellipsoid and on the tangent plane, hence $2a x_0^2+2by_0^2+2cz_0^2+\phi=0 =2+\phi$.
We can now conclude that $T_{x_0}S$ is given by $ a x_0x+ by_0y+ cz_0z-1=0$.
