Whilst There Are Three Characteristic Equations, Only Two of Them Are Linearly Independent? Take the general quasi-linear equation
$$a(x, y, u)u_x + b(x, y, u)u_y - c(x, y, u) = 0. \tag{1}$$
We assume that there exists a solution of the form $u = u(x, y)$.
We can define a solution surface in $(x, y, u)$ space via the implicit form of the solution
$$f(x, y, u) \equiv u(x, y) - u = 0.\tag{2} $$
The normal vector to the solution surface is
$$\nabla f = (f_x, f_y, f_u) = (u_x, u_y, -1).\tag{3}$$
We can therefore write the PDE as the dot product of two vectors
$$au_x + bu_y - c = (a, b, c) \cdot (u_x, u_y, -1) = 0.\tag{4}$$
This shows that the vector $(a(x, y, u), b(x, y, u), c(x, y, u))$ is a tangent vector to the solution surface at the point $(x, y, u)$.
We can construct a curve in $(x, y, u)$ space such that the tangent of the curve is equal to the vector $(a, b, c)$ at every point. Such a curve is called a characteristic curve.
A parameterisation of this curve is given by the equations $x = x(t), y = y(t), u = u(t)$:
$$\left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{du}{dt} \right) = (a, b, c)\tag{5} $$
So we are left with a system of ODEs called the characteristic equations:
$$\frac{dx}{dt} = a(x, y, u), \frac{dy}{dt} = b(x, y, u), \frac{du}{dt} = c(x, y, u). \tag{6}$$
I am told that,

whilst there are three characteristic equations, only two of them are linearly independent. This implies that their solution consists of a two-parameter family of curves in $(x, y, u)$ space.

Why are only two of them linearly independent? And why does this imply that their solution consists of a two-parameter family of curves in $(x, y, u)$ space?
I'm trying to understand this for my upcoming lecture on method of characteristics. Help is much appreciated. 8-)
EDIT: I found the following information in chapter 4.1 of the textbook Essential Partial Differential Equations, by Griffiths, Dold, and Silvester:

We first consider the very special case of (4.1) with $a = b = c = r = 0$, that is
$$pu_x + qu_y = f \ \ \ \text{(4.2)}$$
In many physical applications, see (pde.1), one of the independent variables might represent a time-like variable. In stationary applications both variables might be spatial variables. We begin by considering a curve defined by the height of the surface $z = u(x, y)$ in three dimensions above a path $(x(t), y(t))$ in the $x-y$ plane that is parameterised by $t$. This curve has slope
$$\frac{du}{dt}(x(t), y(t))$$
which, by the chain rule, is given by
$$\frac{du}{dt}(x(t), y(t)) = u_x \frac{dx}{dt} + u_y \frac{dy}{dt} \ \ \ (4.3)$$
Thus, by choosing the parameterization such that
$$\frac{dx}{dt} = p, \frac{dy}{dt} = q, \ \ \ (4.4)$$
the PDE (4.2) reduces to the ODE
$$\frac{du}{dt} = f \ \ \ (4.5)$$
The parameter $t$ is not an intrinsic part of the system and can be avoided by writing the three ODEs in (4.4) and (4.5) in the generic form
$$\frac{dx}{p} = \frac{dy}{q} = \frac{du}{f} \ \ \ (4.6)$$
Paths in the $x-y$ plane described by (4.4) are known as characteristic curves or, simply, as characteristics, and equations (4.6) are known as the characteristic equations fo (4.2). The relations (4.6) define three equations, of which any two are independent.

P.S.: Bolding in textbook quotation is my own.
EDIT 2: Ok, I just realised something. I think this might have something to do with my question:
(From the chapter and textbook described in the first edit.)

Example 4.1
Find the general solution of the PDE $pu_x + u_y = u$, where p is constant. Show that the problem is well posed when solved in the infinite strip { (x, y) : x \in \mathbb{R}, 0 \le y \le Y } and an initial condition $u(x, 0) = g(x), x \in \mathbb{R}$, is applied, where $g$ is a continuous bounded function.
The characteristic equations are
$$\frac{dx}{p} = \frac{dy}{1} = \frac{du}{u}$$
which we may write as
$$\frac{dx}{dy} = p, \frac{du}{dy} = u$$
[...]

See how we got the two ODEs $\frac{dx}{dy} = p, \frac{du}{dy} = u$ from the equation $\frac{dx}{p} = \frac{dy}{1} = \frac{du}{u}$, which has three terms? See how it seems that we cannot get another ODE? If I do the necessary manipulations of $\frac{dx}{dy} = p, \frac{du}{dy} = u$ to get $\frac{dx}{dy} = p$ and $\frac{du}{dy} = u$, if we then attempt to do further manipulations to get a third ODE, then there must be some dependency in the third ODE we get and one of the other two ODEs.
I'm not sure if I'm onto something here, but it's something that I just thought of.
 A: *

*The quasi-linear PDE (1) is invariant under the the following transformation
$$ a^{\prime} ~=~ \lambda a, \qquad b^{\prime} ~=~  \lambda b, \qquad c^{\prime} ~=~  \lambda c, $$
where $a,b,c,\lambda$ are functions of $x,y,u$, and $\lambda$ is nowhere-vanishing. This leads to reparametrization 
$$ \frac{dt}{dt^{\prime}}~=~\lambda  $$
of the 3 ODEs (6) via the chain rule $$\frac{d}{dt^{\prime}}
~=~\frac{dt}{dt^{\prime}}\frac{d}{dt}$$ 
with a new parameter $t^{\prime}$.

*We will make a regularity assumption not mentioned by OP:
$$\text{The 3 functions } (a,b,c) \text{  are continuous and do not vanish simultaneously}. $$
Say that the function $a$ does not vanish in a neighborhood of $(x_0,y_0,u_0)$. (The other 2 cases are similar and left to the reader.) This shows that we in a neighborhood of $(x_0,y_0,u_0)$ can choose $$\lambda~=~a^{-1}, \qquad a^{\prime}~=~ \lambda a~=~1, \qquad t^{\prime}~=~x,$$ so that locally we only have to solve 2 ODEs 
$$ \frac{dy}{dx}~=~\frac{b}{a}, \qquad \frac{dz}{dx}~=~\frac{c}{a}, $$
rather than 3 ODEs (6).
A: Here "only two of them are linearly independent" means we assume $u$ is a function of $x$ and $y$ at the beginning. It follows that the solution of three characteristic equations 
$\frac{dx}{dt}=a(x,y,u),\frac{dy}{dt}=b(x,y,u),\frac{du}{dt}=c(x,y,u)$ 
depends on only two parameters, one of which is $t$ in your notation. It means the solution is 
$x=\alpha(t,s),y=\beta(t,s),u=\gamma(t,s)$
with another parameter, $s$ is used here 
and therefore it is a surface in $(x,y,u)$ space.
All of those comes from the assumption that
$u$ is a function of $x$ and $y$.
The first way to understand it is as follows. This assumption means the manifold $\{(x,y,u)\}$
is 2 dimensional. So if the solution of three characteristic equations is $x=\alpha(t,s_1,s_2), y=\beta(t,s_1,s_2), u=\gamma(t,s_1,s_2)$ with 3 or more independent variables. Then the manifold will have 3 or more dimensions. How could it happen? Think about why the dimension of manifold makess sense.
The second way is by ODE theory, the solution of three characteristic equations is $x=\alpha(t,x_0,y_0,u_0),y=\beta(t,x_0,y_0,u_0),u=\gamma(t,x_0,y_0,u_0)$ with initial data $x_0,y_0,u_0$ at initial time $t_0$. Solving characteristic equations for initial points at a curve which is not parallel to any chracteristic curve $(x_0(s),y_0(s))$, one can get the solution of the original problem as

$x=\alpha(t,x_0(s),y_0(s),u_0(x_0(s),y_0(s))), y=\beta(t,x_0(s),y_0(s),u_0(x_0(s),y_0(s))), u=\gamma(t,x_0(s),y_0(s),u_0(x_0(s),y_0(s)))$

Here, there are only 2 independent variables $t$ and $s$. 
