Which one of the following is the plane containing the line $$\frac{{x - 2}}{2} = \frac{{y - 3}}{3} = \frac{{z - 4}}{5}$$     and parallel to $$z$$-axis ?

Solution :
The equation of the line is $$\frac{{x - 2}}{2} = \frac{{y - 3}}{3} = \frac{{z - 4}}{5} = r$$
Where $$r$$ is a constant. Any point on this line, is given by $$x = 2r + 2,\,y = 3r + 2$$     and $$z = 5r + 4$$
Since, a plane that is parallel to $$z$$-axis will have no $$z$$-co-ordinate, $$z = 0$$
$$z = 0 \Rightarrow 5r + 4 = 0{\text{ or, }}r = \frac{{ - 4}}{5}$$
Putting this value of r for $$x$$ and $$y$$ co-ordinates.
$$\eqalign{ & x = 2r + 2 = 2 \times \left( { - \frac{4}{5}} \right) + 2 \cr & {\text{or, }}5x = - 8 + 10 \cr & {\text{or, }}5x = 2 \cr & x = \frac{2}{5},\,{\text{or }}\,\frac{2}{x} = 5......\left( 1 \right) \cr & {\text{Similarly, }}y = 3r + 3 = 3 \times \left( { - \frac{4}{5}} \right) + 3 \cr & {\text{or, }}5y = - 12 + 15 \cr & {\text{or, }}5y = 3 \cr & y = \frac{3}{5},\,{\text{or }}\frac{3}{y} = 5......\left( 2 \right) \cr & {\text{From equation}}\,\left( 1 \right){\text{ and }}\left( 2 \right) \cr & \frac{2}{x} = \frac{3}{y} \Rightarrow 3x - 2y = 0 \cr} $$