Given $a(t), b(t)$ and $c(t)$ defined on $I \to \mathbb{R}$, we want to define the smooth function, i.e. the envelope, $\gamma(t)$, of the family of lines $a(t)x + b(t)y =c(t)$.
To find $\gamma(t)$ we solve the system
$$ a(t)x + b(t) y = c(t) $$ $$ a'(t)x + b'(t)y = c'(t)$$
If $\begin{vmatrix} a(t) & b(t) \\ a'(t) & b'(t) \end{vmatrix}\neq 0$ then $\gamma(t)$ is unique.
I am now asked to show that the non-singularity and smoothness of $\gamma(t)$ relies on $$ \begin{vmatrix} a(t) & b(t) & c(t) \\ a'(t) & b'(t) & c'(t) \\ a''(t) & b''(t) & c''(t) \end{vmatrix} \neq 0$$
I have absolutely no idea how the determinant of this matrix is related to the above system of equations. Especially since we could differentiate infinitely since $a(t), b(t)$ and $c(t)$ are smooth.
The above matrix looks a lot like the wronskian, so the nonvanishing of the determinant tells us that the three smooth functions are linearly independent. But why does this matter?