The point of contact of the line $x - 2y -1 = 0$ with the parabola
$y^2 = 2 (x - 3),$ is

We have,

   $y^2=2(x-3)\Rightarrow (y-0{)}^2=2(x-3)$            …(i)

as the equation of the parabola. 

The equation of the tangent is

  $x-2y-1=0\Rightarrow y=\frac{x}{2}-\frac{1}{2}\Rightarrow y-0=\frac{1}{2}(x-3)+1$

So, the coordinates of the point of contact are given by

$\begin{array}{l}x-3=\frac{{\displaystyle \frac{1}{2}}}{(1/2{)}^2},,y-0=\frac{2\times 1/2}{1/2}\left[\text{ Using: }x=\frac{a}{m^2}\text{ and }y=\frac{2a}{m}\right]\\ \Rightarrow x=5\text{ and }y=2.\end{array}$