The orthocentre of triangle formed by points $\left(0,0\right)\left(3,4\right)\left(2,3\right)$ is $\left(-\mathrm{k},\mathrm{k}\right)$ then $\mathrm{k}$ is 

The ortho centre of a triangle is the point of concurrence of the altitudes 
The slope of the line joining $\left(0,0\right),\left(2,3\right)$  is    $\mathrm{m}=\frac{3}{2}$
The equation of the altitude from $\left(3,4\right)$ is 

$\begin{array}{c}\mathrm{y}-4=-\frac{2}{3}\left(\mathrm{x}-3\right)\\ 3\mathrm{y}-12=-2\mathrm{x}+6\\ 2\mathrm{x}+3\mathrm{y}-18=0\end{array}$

Point $\left(-\mathrm{k},\mathrm{k}\right)$ lies on the line  $2\mathrm{x}+3\mathrm{y}-18=0$ 
It implies   $-2\mathrm{k}+3\mathrm{k}-18=0\Rightarrow \mathrm{k}=\boxed{18}$