The tangent plane to a smooth ($C^1$) surface $S$ at $a=(a_1,a_2,a_3)$ is the plane of the equations
$$z-a_3=[Df(a_1,a_2)]\begin{bmatrix} x-a_1 \\ y-a_2\end{bmatrix} \\ y-a_2=[Dg(a_1,a_3)]\begin{bmatrix} x-a_1 \\ z-a_3\end{bmatrix} \\x-a_1=[Dh(a_2,a_3)]\begin{bmatrix} y-a_2 \\ z-a_3\end{bmatrix}$$
where $f,g,h:\mathbb R^2 \to \mathbb R$ and $f$ is a function of $(x,y)$ etc. This is a definition in the book "Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach". This is not a unified approach at all. I don't understand where this definition comes from. In linear algebra, a plane in $\mathbb R^3$ is given as follows:
$$\vec n \cdot \vec x = \vec n \cdot \vec p$$
where $\vec n$ is the normal vector and $\vec p$ is the point vector. Alternatively, this equation becomes:
$$\vec x=\vec p + s\vec u + t\vec v$$
in the parametric form.
I can't see how the above equations in the book are related with the definitions of a plane in linear algebra. I have studied linear algebra for the sole purpose of using it when learning multivariable calculus. Now it is frustrating to see that each branch treats subjects differently. So, could you please help me to express a tangent plane to a smooth surface in terms of linear algebra?