Which is the nearest one?

The point of intersection of the given line and the common normal will be the required point.

Common normal will have slope $m=1$ whose equation is obtained as follows,

$\begin{aligned} {y}^2=4(x-10)\\2yy'&=4\\-\dfrac{1}{y'}&=-\dfrac{y}{2}\\m&=-\dfrac{y}{2}\\\Rightarrow (x,y)&=(11,-2)\\\dfrac{y+2}{x-11}&=1\\y&=x-13\end{aligned}$

The required point is obtained by solving the line $x+y=4$ and $x-y=13$ which is $\large \color{#3D99F6}{\boxed{\left(\dfrac{17}{2},-\dfrac{9}{2}\right)}}$ .