Example 1. Given that the following three non-parallel lines intersect at a point, find the value of $\mathbf{k}$;
$$
3 x+2 y+9 k=0,2 x+y-5=0, x+k y+2=0 .
$$
Given that the three lines intersect at a common point,
$$
\therefore\left|\begin{array}{rrr}
3 & 2 & 9 k \\
2 & 1 & -5 \\
1 & k & 2
\end{array}\right|=0,
$$

which simplifies to $3 k^{2}+k-2=0$,
$$
\therefore k_{1}=\frac{2}{3}, k_{2}=-1 \text {. }
$$

When $k=\frac{2}{3}$, the line $3 x+2 y+9 k=0$ coincides with the line $x+k y+2=0$, which is not acceptable, and should be discarded.
$$
\therefore k=-1 \text {. }
$$

From the proof of the sufficient condition, we find that conditions $(\alpha) 、(\beta) 、(\gamma) 、(\delta)$ are all indispensable. Therefore, the sufficient condition for three non-parallel lines to intersect at a common point is that the value of the third-order coefficient determinant is 0. The application value of this sufficient condition lies in its ability to help us solve more complex three-line intersection problems in plane geometry more easily.