If the orthocentre of the triangle whose  vertices  are 
$2\widehat{i}+3\widehat{j}+5\widehat{k},5\widehat{i}+2\widehat{j}+3\widehat{k}$ and $3\widehat{i}+5\widehat{j}+2\widehat{k}$ is $x\widehat{i}+y\widehat{j}+z\widehat{k}$ then 

$$\begin{gathered} \left|AB\right|=\sqrt{9+14}=\sqrt{14} \\ \begin{array}{l}\left|BC\right|=\sqrt{4+9+1}=\sqrt{14}\\ \left|AC\right|=\sqrt{1+4+9}=\sqrt{14}\end{array} \end{gathered}$$
$\because$ an equilateral triangle 
$\therefore$ orthocenter of triangle is coincide with centroid
$\left(\frac{{\displaystyle 2i+3j+5k+5i+2j+3k+3i+5j+2k}}{{\displaystyle 3}}\right)$
$\begin{array}{l}=\frac{15i+15j+15k}{3}=5i+5j+5k\\ x=y=z=5\end{array}$